Kinematics-1D
JEE-Main 2021
Q1
Allen
JEE-Main PYQs
Numerical
A swimmer wants to cross a river from point A to point B. Line AB makes an angle of $30^{\circ}$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $\theta$ with the line AB should be ____°, so that the swimmer reaches point B.
Correct Answer: 30
Explanation:
Ans. (30)
Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them
$\theta = 3 0 ^ {\circ}$
Both velocity vectors are of same magnitude therefore resultant would pass exactly midway through them
$\theta = 3 0 ^ {\circ}$
JEE-Advanced 2019
Q2
Allen
JEE-Advanced PYQs
Numerical
A ball is thrown from ground at an angle $\theta$ with horizontal and with an initial speed $u_{0}$ . For the resulting projectile motion, the magnitude of average velocity of the ball up to the point when it hits the ground for the first time is $V_{1}$ . After hitting the ground, ball rebounds at the same angle $\theta$ but with a reduced speed of $u_{0}/\alpha$ . Its motion continues for a long time as shown in figure. If the magnitude of average velocity of the ball for entire duration of motion is $0.8 V_{1}$ , the value of $\alpha$ is ____.
Correct Answer: 4
Explanation:
Ans. (4)
$\text { Average velocity } = \frac {\text { Total displacement }}{\text { Total time }}$
Total time taken = $t_{1} + t_{2} + t_{3} + \ldots\ldots\ldots$ = $t_{1} + \frac{t_{1}}{\alpha} + \frac{t_{1}}{\alpha^{2}} + \ldots\ldots\ldots$
Total time = $\frac{t_{1}}{1-\frac{1}{\alpha}}$
Total displacement = $v_{1}t_{1} + v_{2}t_{2} +$ .....
$\begin{array}{l} = v _ {1} t _ {1} + \frac {v _ {1}}{\alpha}. \frac {t _ {1}}{\alpha} + \dots \dots \\ = \frac {v _ {1} t _ {1}}{1 - \frac {1}{\alpha^ {2}}} = \frac {v _ {1} t _ {1}}{\left(1 + \frac {1}{\alpha}\right) \left(1 - \frac {1}{\alpha}\right)} \end{array}$
On solving
$\begin{array}{l} \langle v \rangle = \frac {v _ {1} \alpha}{\alpha + 1} = 0. 8 v _ {1} \\ \boxed {\alpha = 4. 0 0} \end{array}$
$\text { Average velocity } = \frac {\text { Total displacement }}{\text { Total time }}$
Total time taken = $t_{1} + t_{2} + t_{3} + \ldots\ldots\ldots$ = $t_{1} + \frac{t_{1}}{\alpha} + \frac{t_{1}}{\alpha^{2}} + \ldots\ldots\ldots$
Total time = $\frac{t_{1}}{1-\frac{1}{\alpha}}$
Total displacement = $v_{1}t_{1} + v_{2}t_{2} +$ .....
$\begin{array}{l} = v _ {1} t _ {1} + \frac {v _ {1}}{\alpha}. \frac {t _ {1}}{\alpha} + \dots \dots \\ = \frac {v _ {1} t _ {1}}{1 - \frac {1}{\alpha^ {2}}} = \frac {v _ {1} t _ {1}}{\left(1 + \frac {1}{\alpha}\right) \left(1 - \frac {1}{\alpha}\right)} \end{array}$
On solving
$\begin{array}{l} \langle v \rangle = \frac {v _ {1} \alpha}{\alpha + 1} = 0. 8 v _ {1} \\ \boxed {\alpha = 4. 0 0} \end{array}$
JEE-Advanced 2018
Q3
Allen
JEE-Advanced PYQs
Numerical
A ball is projected from the ground at an angle of $45^{\circ}$ with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of $30^{\circ}$ with the horizontal surface. The maximum height it reaches after the bounce, in metres, is......
Correct Answer: 30 [29.60, 30.40]
Explanation:
Ans. (30 [29.60, 30.40])
$H _ {1} = \frac {u ^ {2} \sin^ {2} 4 5}{2 g} = 1 2 0$
$\Rightarrow \frac {u ^ {2}}{4 g} = 1 2 0\tag{... (i}$
When half of kinetic energy is lost $v = \frac{u}{\sqrt{2}}$
$H _ {2} = \frac {\left(\frac {u}{\sqrt {2}}\right) ^ {2} \sin^ {2} 3 0}{2 g} = \frac {u ^ {2}}{1 6 g}\tag{... (ii}$
From (i) and (ii)
$H _ {2} = \frac {H _ {1}}{4} = 3 0 m \mathrm{on} 3 0. 0 0$
$H _ {1} = \frac {u ^ {2} \sin^ {2} 4 5}{2 g} = 1 2 0$
$\Rightarrow \frac {u ^ {2}}{4 g} = 1 2 0\tag{... (i}$
When half of kinetic energy is lost $v = \frac{u}{\sqrt{2}}$
$H _ {2} = \frac {\left(\frac {u}{\sqrt {2}}\right) ^ {2} \sin^ {2} 3 0}{2 g} = \frac {u ^ {2}}{1 6 g}\tag{... (ii}$
From (i) and (ii)
$H _ {2} = \frac {H _ {1}}{4} = 3 0 m \mathrm{on} 3 0. 0 0$
JEE-Advanced 2014
Q4
Allen
JEE-Advanced PYQs
Numerical
A rocket is moving in a gravity free space with a constant acceleration of $2 \, ms^{-2}$ along +x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in +x direction with a speed of $0.3 \, ms^{-1}$ relative to the rocket. At the same time, another ball is thrown in -x direction with a speed of $0.2 \, ms^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is
Correct Answer: 8 or 2
Explanation:
Ans. (8 or 2)
Assuming open chamber
$V _ {\mathrm{relative}} = 0. 5 \mathrm{m} / \mathrm{s}$
$S _ {\mathrm{relative}} = 4 m$
$\mathrm{Time} = \frac {4}{0 . 5} = 8 m / s$
Alternate:
Assuming closed chamber
In the frame of chamber :
Maximum displacement of ball A from its left end is $\frac{u_{A}^{2}}{2a} = \frac{(0.3)^{2}}{2(2)} = 0.0225 \, m$
This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be very close to the left end.
Hence, time taken by ball B to reach left end will be given by
$S = u _ {B} t + \frac {1}{2} a t ^ {2}$
$4 = (0. 2) (t) + \frac {1}{2} (2) (t) ^ {2}$
Solving this, we get
$t \approx 2 s$
Assuming open chamber
$V _ {\mathrm{relative}} = 0. 5 \mathrm{m} / \mathrm{s}$
$S _ {\mathrm{relative}} = 4 m$
$\mathrm{Time} = \frac {4}{0 . 5} = 8 m / s$
Alternate:
Assuming closed chamber
In the frame of chamber :
Maximum displacement of ball A from its left end is $\frac{u_{A}^{2}}{2a} = \frac{(0.3)^{2}}{2(2)} = 0.0225 \, m$
This is negligible with respect to the length of chamber i.e. 4m. So, the collision will be very close to the left end.
Hence, time taken by ball B to reach left end will be given by
$S = u _ {B} t + \frac {1}{2} a t ^ {2}$
$4 = (0. 2) (t) + \frac {1}{2} (2) (t) ^ {2}$
Solving this, we get
$t \approx 2 s$
JEE-Advanced 2014
Q5
Allen
JEE-Advanced PYQs
Numerical
Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $100\sqrt{3}$ ms $^{-1}$ . At time t = 0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at $t = t_{0}$ , A just escapes being hit by B, $t_{0}$ in seconds is
Correct Answer: 5
Explanation:
Ans. (5)
As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A
$\begin{array}{l} {V _ {B} \cos 3 0 = 1 0 0 \sqrt {3}} \\ {V _ {B} = 2 0 0} \end{array}$
If A is observer A remains stationary therefore
$t = \frac {5 0 0}{V _ {B} \sin 3 0} = \frac {5 0 0}{1 0 0} = 5$
As observed from A, B moves perpendicular to line of motion of A. It means velocity of B along A is equal to velocity of A
$\begin{array}{l} {V _ {B} \cos 3 0 = 1 0 0 \sqrt {3}} \\ {V _ {B} = 2 0 0} \end{array}$
If A is observer A remains stationary therefore
$t = \frac {5 0 0}{V _ {B} \sin 3 0} = \frac {5 0 0}{1 0 0} = 5$
JEE (Advanced) 2014
Q6
Advance
ARCHIVE: JEE ADVANCED
Numerical
A rocket is moving in a gravity free space with a constant acceleration of $2\mathrm{ms}^{-2}$ along $+x$ direction (shown in figure). The length of a chamber inside the rocket is $4\mathrm{m}$ . A ball is thrown from the left end of the chamber in $+x$ direction with a speed of $0.3\mathrm{ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in $-x$ direction with a speed of $0.2\mathrm{ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each.
Correct Answer: 2 or 8
Explanation:
Let us consider motion of two balls with respect to rocket.
Motion of ball A relative to rocket
Maximum distance of ball A from left wall is
$S _ {A} = \frac {u ^ {2}}{2 a} = \frac {0 . 3 \times 0 . 3}{2 \times 2} = \frac {0 . 0 9}{4} \approx 0. 0 2 \mathrm{m} \left\{\because 0 = u ^ {2} - 2 a S \right\}$
So, collision of two balls will take place very near to left wall.
Motion of ball B relative to rocket
$\begin{array}{c c} \longleftarrow 2 \mathrm{ms} ^ {- 2} \\ \longleftarrow \circ & \longleftarrow + \mathrm{ve} \\ 0. 2 \mathrm{ms} ^ {- 1} & - \mathrm{ve} \end{array}$
$S = u t + \frac {1}{2} a t ^ {2}$
$\Rightarrow - 4 = - 0. 2 t - \left(\frac {1}{2}\right) 2 t ^ {2}$
Solving this equation, we get
$t = 1. 9 \mathrm{s}$
Nearest Integer = 2 s
Motion of ball A relative to rocket
Maximum distance of ball A from left wall is
$S _ {A} = \frac {u ^ {2}}{2 a} = \frac {0 . 3 \times 0 . 3}{2 \times 2} = \frac {0 . 0 9}{4} \approx 0. 0 2 \mathrm{m} \left\{\because 0 = u ^ {2} - 2 a S \right\}$
So, collision of two balls will take place very near to left wall.
Motion of ball B relative to rocket
$\begin{array}{c c} \longleftarrow 2 \mathrm{ms} ^ {- 2} \\ \longleftarrow \circ & \longleftarrow + \mathrm{ve} \\ 0. 2 \mathrm{ms} ^ {- 1} & - \mathrm{ve} \end{array}$
$S = u t + \frac {1}{2} a t ^ {2}$
$\Rightarrow - 4 = - 0. 2 t - \left(\frac {1}{2}\right) 2 t ^ {2}$
Solving this equation, we get
$t = 1. 9 \mathrm{s}$
Nearest Integer = 2 s
JEE (Advanced) 2014
Q7
Advance
ARCHIVE: JEE ADVANCED
Numerical
Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^{\circ}$ and $60^{\circ}$ with respect to the horizontal respectively as shown in figure. The speed of A is $100\sqrt{3}$ ms $^{-1}$ . At time t=0 s, an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at $t=t_{0}$ , A just escapes being hit by B, $t_{0}$ in seconds is
Correct Answer: 5
Explanation:
Relative velocity of B with respect to A is perpendicular to line PA. Therefore, along the line PA, velocity components of A and B should be same.
Q8
Allen
NAT
Numerical
A rocket is fired vertically upwards with initial velocity 40 m/s at the ground level. Its engines then fired and it is accelerated at $2 \, m/s^{2}$ until it reaches an altitude of 1000 m. At that point the engines shut off and the rocket goes into free-fall. If the velocity (in m/s) just before it collides with the ground is $40\alpha$ . Then fill the value of $\alpha$ . Disregard air resistance ( $g = 10m/s^{2}$ ).
Correct Answer: 4
Explanation:
Ans. (4) $v^{2} = u^{2} + 2$ as $v^{2} = 40^{2} + 2(2)(1000) = 5600$ $v^{\prime 2} = v^{2} + 2gs$ $\Rightarrow v^{\prime 2} = 5600 + 2\times 10\times 100$ $\Rightarrow v^{\prime} = 160 = 40\alpha$ $\Rightarrow \alpha = 4$
Q9
Allen
NAT
Numerical
The vertical height y and horizontal distance x of a projectile on a certain planet are given by $x = (3t)m$ , $y = (4t - 6t^{2})m$ where t is in seconds. Find the speed of projection (in m/s).
Correct Answer: 5
Explanation:
Ans. (5) $v_{x} = \frac{dx}{dt} = 3, v_{y} = \frac{dy}{dt} = 4 - 12t$ At $t = 0, \vec{u} = u_{x}\hat{i} + u_{y}\hat{j} = 3\hat{i} + 4\hat{j}$ Speed of projection $= u = \sqrt{3^2 + 4^2} = 5m / s$
Q10
Allen
NAT
Numerical
A particle is thrown with a speed $60 \, ms^{-1}$ at an angle $60^{\circ}$ to the horizontal. When the particle makes an angle $30^{\circ}$ with the horizontal in downward direction, it's speed at that instant is v. What is the value of $v^{2}$ in SI units?
Correct Answer: 1200
Explanation:
Ans. (1200) Horizontal velocity remain constant. $60\cos 60^{\circ} = v\cos 30^{\circ}$ $v = 20\sqrt{3}$ $v^{2} = 1200$
Q11
Allen
NAT
Numerical
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high (in m) above the ground can the cricketer throw the same ball?
Correct Answer: 50
Explanation:
Ans. (50)
$H _ {\mathrm{max}} = u ^ {2} \sin^ {2} \theta / 2 g = u ^ {2} / 2 g$
$R _ {\max} = \left(u ^ {2} \sin 2 \theta / g\right) _ {\max} = u ^ {2} / g$
$2 H _ {\mathrm{max}} = \mathrm{Range}$
$2 H _ {\mathrm{max}} = 1 0 0$
$H _ {\mathrm{max}} = 5 0 m$
$H _ {\mathrm{max}} = u ^ {2} \sin^ {2} \theta / 2 g = u ^ {2} / 2 g$
$R _ {\max} = \left(u ^ {2} \sin 2 \theta / g\right) _ {\max} = u ^ {2} / g$
$2 H _ {\mathrm{max}} = \mathrm{Range}$
$2 H _ {\mathrm{max}} = 1 0 0$
$H _ {\mathrm{max}} = 5 0 m$
Q12
Allen
NAT
Numerical
A particle is projected upwards with a velocity of 100 m/s at an angle of $60^{\circ}$ with the vertical. Find the time (in sec) when the particle will move perpendicular to its initial direction, taking $g = 10 \, m/s^{2}$ .
Correct Answer: 20
Explanation:
Ans. (20) $U = 100\cos 30\hat{t} +100\sin 30^{\circ}\hat{j}$ $\vec{V} = 10\cos 30\hat{t} +(100\sin 30 - 10t)\hat{j}$ $\vec{U}.\vec{V} = 0$ $\Rightarrow \frac{100}{10\times\frac{1}{2}} = 20\sec$
Q13
Allen
NAT
Numerical
A ball is thrown horizontally from a cliff such that it strikes ground after 5 s. The line of sight from the point of projection to the point of landing makes an angle of $37^{\circ}$ with the horizontal. If the initial velocity of projection is $\frac{50x}{3}$ . Find x.
Correct Answer: 2
Explanation:
Ans. (2) $t^2 = \frac{2h}{g} = 25$ $\Rightarrow h = 125m$ $x = u_{x}\times t$ $h\cot 37^{\circ} = u_{x}\times 5$ $u_{x} = 100 / 3$
Q14
Allen
NAT
Numerical
A ball is dropped from rest from a tower of height 5m. As a result of the wind it lands at a distance 6m from the bottom of the tower as shown. Assuming no air resistance but that the wind gives the ball a constant horizontal velocity v. Find value of v in m/s.
Correct Answer: 6
Explanation:
Ans. (6)
$\begin{array}{l} {x = \sqrt {\frac {2 h}{g}} v} \\ {\Rightarrow \quad v = \frac {x}{\sqrt {\frac {2 h}{g}}} = \frac {6}{\sqrt {\frac {2 \times 5}{1 0}}} = 6 m / s} \end{array}$
$\begin{array}{l} {x = \sqrt {\frac {2 h}{g}} v} \\ {\Rightarrow \quad v = \frac {x}{\sqrt {\frac {2 h}{g}}} = \frac {6}{\sqrt {\frac {2 \times 5}{1 0}}} = 6 m / s} \end{array}$
Q15
Allen
NAT
Numerical
A cuboidal elevator cabin is shown in the figure. A ball is thrown from point A on the floor of cabin when the elevator is falling under gravity. The plane of motion is ABCD and the angle of projection of the ball with AB, relative to elevator, if the ball collides with point C, is $\alpha$ . Then find the value of tan $\alpha$ .
Correct Answer: 5
Explanation:
Ans. (5)
$\begin{array}{l} \tan \alpha = \frac {A D}{A B} \\ \frac {2 . 5}{0 . 5} \\ \tan \alpha = 5 \end{array}$
$\begin{array}{l} \tan \alpha = \frac {A D}{A B} \\ \frac {2 . 5}{0 . 5} \\ \tan \alpha = 5 \end{array}$
Q16
Allen
NAT
Numerical
Two particles P and Q are launched simultaneously as shown in figure. Find the minimum distance between particles in meters.
Correct Answer: 6
Explanation:
Ans. (6)
$\begin{array}{r l} & {\vec {v} _ {P Q} = \vec {v} _ {P} - \vec {v} _ {Q}} \\ & {\qquad = \left(\frac {1 0}{\sqrt {2}} \hat {\imath} + \frac {1 0}{\sqrt {2}} \hat {j}\right) - \left(- \frac {7 0}{\sqrt {2}} \hat {\imath} + \frac {7 0}{\sqrt {2}} \hat {j}\right)} \\ & {\qquad = 4 0 \sqrt {2} \hat {\imath} - 3 0 \sqrt {2} \hat {j}} \\ & {A B = 9 0 \times \frac {4}{3} = 1 2 0 m} \\ & {\Rightarrow \quad Q B = 1 0 m} \\ & {\Rightarrow \quad Q M = 1 0 \sin 3 7 ^ {\circ} = 1 0 \times \frac {3}{5} = 6 m} \end{array}$
$\begin{array}{r l} & {\vec {v} _ {P Q} = \vec {v} _ {P} - \vec {v} _ {Q}} \\ & {\qquad = \left(\frac {1 0}{\sqrt {2}} \hat {\imath} + \frac {1 0}{\sqrt {2}} \hat {j}\right) - \left(- \frac {7 0}{\sqrt {2}} \hat {\imath} + \frac {7 0}{\sqrt {2}} \hat {j}\right)} \\ & {\qquad = 4 0 \sqrt {2} \hat {\imath} - 3 0 \sqrt {2} \hat {j}} \\ & {A B = 9 0 \times \frac {4}{3} = 1 2 0 m} \\ & {\Rightarrow \quad Q B = 1 0 m} \\ & {\Rightarrow \quad Q M = 1 0 \sin 3 7 ^ {\circ} = 1 0 \times \frac {3}{5} = 6 m} \end{array}$
Q17
Allen
NAT
Numerical
Velocity of the boat with respect to river is 10 m/s. From point A it is steered in the direction shown. Find the distance (in m) it will reach on the opposite bank from point B.
Correct Answer: 30
Explanation:
Ans. (30)
$\begin{array}{l} {t = \frac {1 2 0}{8} = 1 5 \sec} \\ {\Delta = (6 - 4) \times 1 5 = 3 0 m} \end{array}$
$\begin{array}{l} {t = \frac {1 2 0}{8} = 1 5 \sec} \\ {\Delta = (6 - 4) \times 1 5 = 3 0 m} \end{array}$
Q18
Allen
JEE-Advanced Pattern
Numerical
A body starting from rest accelerates uniformly at the rate of $10 \, cm s^{-2}$ and retards uniformly at the rate of $20 \, cm/sec^{2}$ . Find the least time in which it can complete a journey of 5 km if the maximum velocity attained by the body is $72 \, km/h^{-1}$ .
Correct Answer: 400
Explanation:
Ans. (400) $u = 0, \alpha = 10cm / s^2, \beta = 20cm / s^2$ $5km = \frac{1}{2} \times t \times v_m$ $t = \alpha v_0 + \beta v_0$
Q19
Allen
JEE-Advanced Pattern
Numerical
A ball is thrown vertically upwards from the ground. It crosses a point at the height of 25 m twice at an interval of 4 secs. The ball was thrown with the velocity of
Correct Answer: 30
Explanation:
Ans. (30)
$V = 10(2)$ $= 20m / s$ $v^{\prime 2} = 20^{2} + 2\times 10\times 25$ $= 400 + 500$ $= 900$ $v^{\prime} = 30m / s$ Ans.
$V = 10(2)$ $= 20m / s$ $v^{\prime 2} = 20^{2} + 2\times 10\times 25$ $= 400 + 500$ $= 900$ $v^{\prime} = 30m / s$ Ans.
Q20
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
If the body covers equal displacements in successive intervals of time $t_{1}$ , $t_{2}$ and $t_{3}$ then show that $\frac{1}{t_{1}} - \frac{1}{t_{2}} + \frac{1}{t_{3}} = \frac{k}{t_{1} + t_{2} + t_{3}}$ . Find k.
Correct Answer: 3
Explanation:
Let velocities at A, B, C and D be $v_{A}$ , $v_{B}$ , $v_{C}$ and $v_{D}$ respectively.
$\begin{array}{c c c c c c c} & I & & I & & I \\ \hline A & t _ {1} & B & t _ {2} & C & t _ {3} & D \end{array}$
For AB
$l = \left(\frac {v _ {A} + v _ {B}}{2}\right) t _ {1}\tag{... (1}$
For BC
For the interval BC:
$l = \left(\frac{v_B + v_C}{2}\right) t_2 \tag{2}$
For the interval CD:
$l = \left(\frac{v_C + v_D}{2}\right) t_3 \tag{3}$
Combining equations (1), (2), and (3):
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l}(v_A + v_B - v_B - v_C + v_C + v_D)$
$\implies \frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l}(v_A + v_D) \tag{4}$
For the entire interval AD:
$3l = \left(\frac{v_A + v_D}{2}\right) (t_1 + t_2 + t_3)$
$\implies \frac{1}{2l}(v_A + v_D) = \frac{3}{t_1 + t_2 + t_3} \tag{5}$
Finally, substituting (5) into (4):
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{3}{t_1 + t_2 + t_3}$
$\implies k = 3$
$\begin{array}{c c c c c c c} & I & & I & & I \\ \hline A & t _ {1} & B & t _ {2} & C & t _ {3} & D \end{array}$
For AB
$l = \left(\frac {v _ {A} + v _ {B}}{2}\right) t _ {1}\tag{... (1}$
For BC
For the interval BC:
$l = \left(\frac{v_B + v_C}{2}\right) t_2 \tag{2}$
For the interval CD:
$l = \left(\frac{v_C + v_D}{2}\right) t_3 \tag{3}$
Combining equations (1), (2), and (3):
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l}(v_A + v_B - v_B - v_C + v_C + v_D)$
$\implies \frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{1}{2l}(v_A + v_D) \tag{4}$
For the entire interval AD:
$3l = \left(\frac{v_A + v_D}{2}\right) (t_1 + t_2 + t_3)$
$\implies \frac{1}{2l}(v_A + v_D) = \frac{3}{t_1 + t_2 + t_3} \tag{5}$
Finally, substituting (5) into (4):
$\frac{1}{t_1} - \frac{1}{t_2} + \frac{1}{t_3} = \frac{3}{t_1 + t_2 + t_3}$
$\implies k = 3$
Q21
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Between two stations a train accelerates uniformly at first, then moves with a constant speed and finally retards uniformly. If the ratios of the time taken are 1:8:1 and the greatest speed attained by the train is $60 \, kmh^{-1}$ , find the average speed, in $ms^{-1}$ , over the whole journey.
Correct Answer: 15
Explanation:
Since time to accelerate from zero to $60 \, kmh^{-1}$ is equal to the time taken to decelerate from $60 \, kmh^{-1}$ to zero. Hence, we can say both OA and BC are identical intervals. So, total distance travelled from O to A to B to C is
$l = l _ {1} + l _ {2} + l _ {1} = \frac {1}{2} v _ {\mathrm{max}} t + (v _ {\mathrm{max}}) 8 t + \frac {1}{2} v _ {\mathrm{max}} t$
Since,
Average Speed= $\frac{Total Distance Travelled}{Total Time Taken}$
$\Rightarrow v _ {a v} = \frac {l _ {1} + l _ {2} + l _ {1}}{t + 8 t + t}$
$\begin{array}{l l} \Rightarrow & v _ {a v} = \frac {\frac {1}{2} v _ {\max} t + (v _ {\max}) 8 t + \frac {1}{2} v _ {\max} t}{1 0 t} \\ \Rightarrow & v _ {a v} = \frac {9}{1 0} v _ {\max} = \frac {9}{1 0} (6 0 \mathrm{kmh} ^ {- 1}) = 5 4 \mathrm{kmh} ^ {- 1} \\ \Rightarrow & v _ {a v} = 5 4 \times \frac {5}{1 8} \mathrm{ms} ^ {- 1} = 1 5 \mathrm{ms} ^ {- 1} \end{array}$
$\begin{array}{l l} 3. & u = 2 7 \mathrm{ms} ^ {- 1} \text {at} t _ {0} = 0 \mathrm{s}. \text {Since} \\ & d v = a d t \\ & \Rightarrow \int_ {2 7} ^ {v} d v = \int_ {0} ^ {t} - 6 t d t \\ & \Rightarrow v = (2 7 - 3 t ^ {2}) \mathrm{ms} ^ {- 1} \\ & \text {At} v = 0, \text {from equation ... (1)} \\ & 0 = 2 7 - 3 t ^ {2} \\ & \Rightarrow t = 3 \mathrm{s} \\ & \text {So, it will stop at} t = 3 \mathrm{s} \\ & \text {Since} d x = v d t \\ & \Rightarrow \int_ {0} ^ {x} d x = \int_ {0} ^ {3} (2 7 - 3 t ^ {2}) d t \\ & \Rightarrow x = (2 7 t - t ^ {3}) | _ {0} ^ {3} = 2 7 (3) - (3) ^ {3} = 5 4 \mathrm{m} \end{array}\tag{... (1}$
$l = l _ {1} + l _ {2} + l _ {1} = \frac {1}{2} v _ {\mathrm{max}} t + (v _ {\mathrm{max}}) 8 t + \frac {1}{2} v _ {\mathrm{max}} t$
Since,
Average Speed= $\frac{Total Distance Travelled}{Total Time Taken}$
$\Rightarrow v _ {a v} = \frac {l _ {1} + l _ {2} + l _ {1}}{t + 8 t + t}$
$\begin{array}{l l} \Rightarrow & v _ {a v} = \frac {\frac {1}{2} v _ {\max} t + (v _ {\max}) 8 t + \frac {1}{2} v _ {\max} t}{1 0 t} \\ \Rightarrow & v _ {a v} = \frac {9}{1 0} v _ {\max} = \frac {9}{1 0} (6 0 \mathrm{kmh} ^ {- 1}) = 5 4 \mathrm{kmh} ^ {- 1} \\ \Rightarrow & v _ {a v} = 5 4 \times \frac {5}{1 8} \mathrm{ms} ^ {- 1} = 1 5 \mathrm{ms} ^ {- 1} \end{array}$
$\begin{array}{l l} 3. & u = 2 7 \mathrm{ms} ^ {- 1} \text {at} t _ {0} = 0 \mathrm{s}. \text {Since} \\ & d v = a d t \\ & \Rightarrow \int_ {2 7} ^ {v} d v = \int_ {0} ^ {t} - 6 t d t \\ & \Rightarrow v = (2 7 - 3 t ^ {2}) \mathrm{ms} ^ {- 1} \\ & \text {At} v = 0, \text {from equation ... (1)} \\ & 0 = 2 7 - 3 t ^ {2} \\ & \Rightarrow t = 3 \mathrm{s} \\ & \text {So, it will stop at} t = 3 \mathrm{s} \\ & \text {Since} d x = v d t \\ & \Rightarrow \int_ {0} ^ {x} d x = \int_ {0} ^ {3} (2 7 - 3 t ^ {2}) d t \\ & \Rightarrow x = (2 7 t - t ^ {3}) | _ {0} ^ {3} = 2 7 (3) - (3) ^ {3} = 5 4 \mathrm{m} \end{array}\tag{... (1}$
Q22
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A sphere is fired downwards into a medium with an initial speed of $27 \, ms^{-1}$ . If it experiences a deceleration of $a = (-6t) \, \text{ms}^{-2}$ , where t is in seconds, determine the distance, in metre, travelled before it stops.
Correct Answer: 54
Explanation:
54
Q23
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle travels along a straight line such that in 2 s it moves from an initial position $x_{A} = +0.5 \, \text{m}$ to a position $x_{B} = -1.5 \, \text{m}$ . Then in another 4 s it moves from $x_{B}$ to $x_{C} = +2.5 \, \text{m}$ . Determine the particle's average speed, in $\text{ms}^{-1}$ , during the 6 s time interval.
Correct Answer: 1
Explanation:
$x_{\mathrm{total}} = (0.5 + 1.5 + 1.5 + 2.5) = 6 \mathrm{~m}$
$t = (2 + 4) = 6 \mathrm{s}$
$\left(v _ {s p}\right) _ {a v g} = \frac {x _ {\mathrm{total}}}{t} = \frac {6}{6} = 1 \mathrm{ms} ^ {- 1}$
$t = (2 + 4) = 6 \mathrm{s}$
$\left(v _ {s p}\right) _ {a v g} = \frac {x _ {\mathrm{total}}}{t} = \frac {6}{6} = 1 \mathrm{ms} ^ {- 1}$
Q24
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle moving with constant acceleration along a straight line covers the distance between two points 80 m apart in 10 s. Its speed as at passes second point is $18 \, ms^{-1}$ .
(A) What is its speed, in ms $^{-1}$ , at the first point?
(B) What is its acceleration in ms $^{-2}$ ?
(C) At what prior distance, in m and time, in s from first point, the particle reverses its direction of motion?
(D) What is the total distance travelled, in m, during this 10 s?
(A) What is its speed, in ms $^{-1}$ , at the first point?
(B) What is its acceleration in ms $^{-2}$ ?
(C) At what prior distance, in m and time, in s from first point, the particle reverses its direction of motion?
(D) What is the total distance travelled, in m, during this 10 s?
Correct Answer: (a) 2 (b) 2 (c) 1, 1 (d) 82
Explanation:
$\begin{array}{l l} \text {(a)} & 8 0 = \left(\frac {u + 1 8}{2}\right) 1 0 \\ & \Rightarrow \quad 1 6 = u + 1 8 \\ & \xrightarrow {\text {1 m}} \\ & t = 1 \xrightarrow {\text {u = 2 ms} ^ {- 1}} t = 0 \xrightarrow {\text {a}} \xrightarrow {\oplus} \\ & \xrightarrow {\text {1 m}} \xrightarrow {\text {80 m}} \\ & \Rightarrow \quad u = - 2 \mathrm{ms} ^ {- 1} \\ \text {(b)} & 1 8 = - 2 + a (1 0) \\ & \Rightarrow \quad a = 2 \mathrm{ms} ^ {- 2} \\ \text {(c)} & v ^ {2} - u ^ {2} = 2 a s \\ & \Rightarrow \quad 0 ^ {2} - (2) ^ {2} = 2 (- 2) s \\ & \Rightarrow \quad s = 1 \mathrm{m} \\ & \qquad \qquad \qquad \qquad \qquad 0 = - 2 + (2) t \\ & \Rightarrow \quad t = 1 \mathrm{s} \\ \text {(d)} & s _ {\text {total}} = 8 0 + | - 1 | + | 1 | = 8 2 \mathrm{m} \end{array}$
Q25
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Tests reveal that a normal driver takes about 0.75 s before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1% alcohol in his system to do the same. If such drivers are travelling on a straight road at $44 \, ms^{-1}$ and their cars can decelerate at $2 \, ms^{-2}$ , determine the shortest stopping distance (d) for each, in metre, from the moment they see the pedestrians.
← d →
← d →
Correct Answer: 517, 616
Explanation:
For normal driver, the car moves a distance of $x_{0}=vt=44(0.75)=33\ m$ before he or she reacts and decelerates the car. The stopping distance can be obtained using
$\begin{array}{r l} & v ^ {2} - u ^ {2} = 2 a (x - x _ {0}), \\ \text {where} & x = ?, x _ {0} = 3 3 \mathrm{m}, a = - 2 \mathrm{ms} ^ {- 2}, a = 4 4 \mathrm{ms} ^ {- 1} \\ \Rightarrow & 0 ^ {2} - 4 4 ^ {2} = 2 (- 2) (x - 3 3) \\ \Rightarrow & x = 5 1 7 \mathrm{m} \end{array}$
For a drunk driver, the car moves a distance of $x_{0}^{\prime}=vt=44(3)=132\ m$ before he or she reacts and decelerates the car. The stopping distance can be obtained again by using
$v ^ {2} - u ^ {2} = 2 a \left(x ^ {\prime} - x _ {0} ^ {\prime}\right)$
$\begin{array}{r l} \Rightarrow & 0 ^ {2} - 4 4 ^ {2} = 2 (- 2) (x ^ {\prime} - 1 3 2) \\ & x ^ {\prime} = 6 1 6 \mathrm{m} \end{array}$
$\begin{array}{r l} & v ^ {2} - u ^ {2} = 2 a (x - x _ {0}), \\ \text {where} & x = ?, x _ {0} = 3 3 \mathrm{m}, a = - 2 \mathrm{ms} ^ {- 2}, a = 4 4 \mathrm{ms} ^ {- 1} \\ \Rightarrow & 0 ^ {2} - 4 4 ^ {2} = 2 (- 2) (x - 3 3) \\ \Rightarrow & x = 5 1 7 \mathrm{m} \end{array}$
For a drunk driver, the car moves a distance of $x_{0}^{\prime}=vt=44(3)=132\ m$ before he or she reacts and decelerates the car. The stopping distance can be obtained again by using
$v ^ {2} - u ^ {2} = 2 a \left(x ^ {\prime} - x _ {0} ^ {\prime}\right)$
$\begin{array}{r l} \Rightarrow & 0 ^ {2} - 4 4 ^ {2} = 2 (- 2) (x ^ {\prime} - 1 3 2) \\ & x ^ {\prime} = 6 1 6 \mathrm{m} \end{array}$