Q551
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The driver of the train A moving with a speed of $144 \, kmh^{-1}$ sights another train B 1 km ahead of him. The train B is moving with a uniform speed of $108 \, kmh^{-1}$ . The driver of the train A immediately applies brakes producing a constant retardation and just manages to avoid a collision. What is the retardation of the train A, in $cms^{-2}$ ? For how long is this retardation produced?
Correct Answer: 5, 200
Explanation:
Initial relative velocity of approach of A and B is
$u _ {r} = (1 4 4 - 1 0 8) \mathrm{kmh} ^ {- 1} = 3 6 \mathrm{kmh} ^ {- 1} = 1 0 \mathrm{ms} ^ {- 1}$
The driver of train A applies brakes, so if a is the retardation, then retardation of A w.r.t. B is $a_{r} = -a$ . With this retardation he just manages to avoid the collision, so
$\begin{array}{r l} & v _ {r} ^ {2} - u _ {r} ^ {2} = 2 a _ {r} l _ {r} \\ \Rightarrow & O ^ {2} - (1 0) ^ {2} = 2 (- a) (1 0 0 0) \\ \Rightarrow & a = \frac {1 0 0}{2 0 0 0} = \frac {1}{2 0} \mathrm{ms} ^ {- 2} = 5 \mathrm{cms} ^ {- 2} \end{array}$
If this retardation is produced for a duration on t (say), then
$\begin{array}{r l} & v _ {r} = u _ {r} + a _ {r} t \\ \Rightarrow & 0 = (1 0) - \left(\frac {1}{2 0}\right) t \\ \Rightarrow & t = 2 0 0 \mathrm{s} \end{array}$
$u _ {r} = (1 4 4 - 1 0 8) \mathrm{kmh} ^ {- 1} = 3 6 \mathrm{kmh} ^ {- 1} = 1 0 \mathrm{ms} ^ {- 1}$
The driver of train A applies brakes, so if a is the retardation, then retardation of A w.r.t. B is $a_{r} = -a$ . With this retardation he just manages to avoid the collision, so
$\begin{array}{r l} & v _ {r} ^ {2} - u _ {r} ^ {2} = 2 a _ {r} l _ {r} \\ \Rightarrow & O ^ {2} - (1 0) ^ {2} = 2 (- a) (1 0 0 0) \\ \Rightarrow & a = \frac {1 0 0}{2 0 0 0} = \frac {1}{2 0} \mathrm{ms} ^ {- 2} = 5 \mathrm{cms} ^ {- 2} \end{array}$
If this retardation is produced for a duration on t (say), then
$\begin{array}{r l} & v _ {r} = u _ {r} + a _ {r} t \\ \Rightarrow & 0 = (1 0) - \left(\frac {1}{2 0}\right) t \\ \Rightarrow & t = 2 0 0 \mathrm{s} \end{array}$
Q552
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A sailor in a boat, which is going due east with a speed of $8 \, ms^{-1}$ observes that a submarine is heading towards north at a speed of $12 \, ms^{-1}$ and sinking at a rate of $2 \, ms^{-1}$ . The commander of submarine observes a helicopter ascending at a rate of $5 \, ms^{-1}$ and heading towards west with $4 \, ms^{-1}$ . Find the actual speed of the helicopter and its speed with respect to boat, both in $ms^{-1}$ .
Correct Answer: 13, 13
Explanation:
For convenience, let us take $\hat{i}$ along east, $\hat{j}$ along north and $\hat{k}$ vertically downwards.
$\vec {v} _ {b} = 8 \hat {i}, \vec {v} _ {s b} = 1 2 \hat {j} + 2 \hat {k}$
So, absolute velocity of submarine is
$\vec {v} _ {s} = \vec {v} _ {s b} + \vec {v} _ {b} = 8 \hat {i} + 1 2 \hat {j} + 2 \hat {k} \qquad \left\{\because \vec {v} _ {s b} = \vec {v} _ {s} - \vec {v} _ {b} \right\}$
$\begin{array}{r l} \Rightarrow & \vec {v} _ {h} = \vec {v} _ {h s} + \vec {v} _ {s} = 4 \hat {i} + 1 2 \hat {j} - 3 \hat {k} \qquad \left\{\because \vec {v} _ {h s} = \vec {v} _ {h} - \vec {v} _ {s} \right\} \\ \Rightarrow & | \vec {v} _ {h} | = \sqrt {(4) ^ {2} + (1 2) ^ {2} + (3) ^ {2}} = 1 3 \mathrm{ms} ^ {- 1} \end{array}$
Since, $\vec{v}_{hb} = \vec{v}_h - \vec{v}_b = -4\hat{i} +12\hat{j} -3\hat{k}$
$\Rightarrow \left| \vec {v} _ {h b} \right| = \sqrt {(- 4) ^ {2} + (1 2) ^ {2} + (- 3) ^ {2}} = 1 3 \mathrm{ms} ^ {- 1}$
$\vec {v} _ {b} = 8 \hat {i}, \vec {v} _ {s b} = 1 2 \hat {j} + 2 \hat {k}$
So, absolute velocity of submarine is
$\vec {v} _ {s} = \vec {v} _ {s b} + \vec {v} _ {b} = 8 \hat {i} + 1 2 \hat {j} + 2 \hat {k} \qquad \left\{\because \vec {v} _ {s b} = \vec {v} _ {s} - \vec {v} _ {b} \right\}$
$\begin{array}{r l} \Rightarrow & \vec {v} _ {h} = \vec {v} _ {h s} + \vec {v} _ {s} = 4 \hat {i} + 1 2 \hat {j} - 3 \hat {k} \qquad \left\{\because \vec {v} _ {h s} = \vec {v} _ {h} - \vec {v} _ {s} \right\} \\ \Rightarrow & | \vec {v} _ {h} | = \sqrt {(4) ^ {2} + (1 2) ^ {2} + (3) ^ {2}} = 1 3 \mathrm{ms} ^ {- 1} \end{array}$
Since, $\vec{v}_{hb} = \vec{v}_h - \vec{v}_b = -4\hat{i} +12\hat{j} -3\hat{k}$
$\Rightarrow \left| \vec {v} _ {h b} \right| = \sqrt {(- 4) ^ {2} + (1 2) ^ {2} + (- 3) ^ {2}} = 1 3 \mathrm{ms} ^ {- 1}$
Q553
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The speed-time graph of a particle moving along a fixed direction is shown in figure. Obtain the distance travelled by the particle, in metre and the average speed of the particle in $ms^{-1}$ between
(A) $t = 0$ to $10\mathrm{s}$ (b t = 2 to 6 s. 
Correct Answer: (a) 60, 6 (b) 10.8, 9
Explanation:
(a) Distance travelled by the particle between t = 0 to 10 s is given by
$\begin{array}{l l} & s = \text { Area of } \Delta O A B = \frac {1}{2} \times 1 0 \times 1 2 = 6 0 \mathrm{m} \\ & \text { Average speed } = \frac {\text { total distance covered }}{\text { total time taken }} \\ \Rightarrow & \text { Average speed } = \frac {6 0}{1 0} = 6 \mathrm{ms} ^ {- 1} \end{array}$
(b) Acceleration of the particle during journey OA is given by
$v = u + at$ where $v = 12\mathrm{ms}^{-1}$ , $u = 0$ and $t = 5\mathrm{s}$
$\Rightarrow a = + 2. 4 \mathrm{ms} ^ {- 2}$
Similarly, acceleration of the particle during journey AB is given by
$v = u + at$ where $v = 0$ , $u = 12 \mathrm{~ms}^{-1}$ and $t = 5 \mathrm{~s}$
$\Rightarrow a = - 2. 4 \mathrm{ms} ^ {- 2}$
Velocity of the particle after 2 s from the start is given by
$v = u + a t = 0 + 2. 4 \times 2 = 4. 8 \mathrm{ms} ^ {- 1}$
So, distance covered by the particle between t = 2 to 5 s (in 3 s) is given by
$\begin{array}{r l} & s _ {1} = u t + \frac {1}{2} a t ^ {2} \\ \Rightarrow & s _ {1} = (4. 8) (3) + \frac {1}{2} (2. 4) (3) ^ {2} = 2 5. 2 \mathrm{m} \end{array}$
Distance covered by the particle from t = 5 s to 6 s (in 1 s) is given by
$\begin{array}{r l} & s _ {2} = u t + \frac {1}{2} a t ^ {2} \\ \Rightarrow & s _ {2} = (1 2) (1) + \frac {1}{2} (- 2. 4) (1) ^ {2} = 1 0. 8 \mathrm{m} \end{array}$
Total distance travelled from t = 2 s to 6 s is
$s = s _ {1} + s _ {2} = 2 5. 2 + 1 0. 8 = 3 6 \mathrm{m}$
Average speed in the interval t = 2 s to 6 s
$v _ {a v} = \frac {\mathrm{totaldistancecovered}}{\mathrm{totaltimetaken}} = \frac {3 6}{4} = 9 \mathrm{ms} ^ {- 1}$
$\begin{array}{l l} & s = \text { Area of } \Delta O A B = \frac {1}{2} \times 1 0 \times 1 2 = 6 0 \mathrm{m} \\ & \text { Average speed } = \frac {\text { total distance covered }}{\text { total time taken }} \\ \Rightarrow & \text { Average speed } = \frac {6 0}{1 0} = 6 \mathrm{ms} ^ {- 1} \end{array}$
(b) Acceleration of the particle during journey OA is given by
$v = u + at$ where $v = 12\mathrm{ms}^{-1}$ , $u = 0$ and $t = 5\mathrm{s}$
$\Rightarrow a = + 2. 4 \mathrm{ms} ^ {- 2}$
Similarly, acceleration of the particle during journey AB is given by
$v = u + at$ where $v = 0$ , $u = 12 \mathrm{~ms}^{-1}$ and $t = 5 \mathrm{~s}$
$\Rightarrow a = - 2. 4 \mathrm{ms} ^ {- 2}$
Velocity of the particle after 2 s from the start is given by
$v = u + a t = 0 + 2. 4 \times 2 = 4. 8 \mathrm{ms} ^ {- 1}$
So, distance covered by the particle between t = 2 to 5 s (in 3 s) is given by
$\begin{array}{r l} & s _ {1} = u t + \frac {1}{2} a t ^ {2} \\ \Rightarrow & s _ {1} = (4. 8) (3) + \frac {1}{2} (2. 4) (3) ^ {2} = 2 5. 2 \mathrm{m} \end{array}$
Distance covered by the particle from t = 5 s to 6 s (in 1 s) is given by
$\begin{array}{r l} & s _ {2} = u t + \frac {1}{2} a t ^ {2} \\ \Rightarrow & s _ {2} = (1 2) (1) + \frac {1}{2} (- 2. 4) (1) ^ {2} = 1 0. 8 \mathrm{m} \end{array}$
Total distance travelled from t = 2 s to 6 s is
$s = s _ {1} + s _ {2} = 2 5. 2 + 1 0. 8 = 3 6 \mathrm{m}$
Average speed in the interval t = 2 s to 6 s
$v _ {a v} = \frac {\mathrm{totaldistancecovered}}{\mathrm{totaltimetaken}} = \frac {3 6}{4} = 9 \mathrm{ms} ^ {- 1}$
Q554
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
At the moment t = 0, a particle leaves the origin and moves in the positive direction of the x-axis. Its velocity varies with time as $v = v_{0} \left(1 - \frac{t}{5}\right)$ where $v_{0} = 10 \, cms^{-1}$ is the initial speed of the particle. The particle will be at the distance of 10 cm from the origin at three different instants. Find out the approximate time interval between the second and the third instant.
Correct Answer: 2
Explanation:
$\frac{dx}{dt} = v_0 - \frac{v_0t}{5}$
$\Rightarrow x = v _ {0} t - \frac {v _ {0} t ^ {2}}{1 0}$
where x can be either +10 cm or -10 cm
For $x = +10 \, \mathrm{cm}$ , we have $10t - t^2 = 10$
$\Rightarrow \quad t = \left(\frac {1 0 \pm \sqrt {6 0}}{2}\right) s$
Hence we have
$t _ {1} = \left(\frac {1 0 - \sqrt {6 0}}{2}\right) \mathrm{s}$
{First Instant}
$\Rightarrow t _ {2} = \left(\frac {1 0 + \sqrt {6 0}}{2}\right) \mathrm{s}$
{Second Instant}
For $x = -10\mathrm{cm}$ , we have $10t - t^2 = -10$
$\Rightarrow t = \left(\frac {1 0 \pm \sqrt {1 4 0}}{2}\right) s$
$\Rightarrow t _ {3} = \left(\frac {1 0 + \sqrt {1 4 0}}{2}\right) \mathrm{s}$
{Third Instant}
However, we will not get fourth instant as time cannot be negative.
Time interval between the second and the third instant is $\Delta t = t_{3} - t_{2} \approx 2$ s
$\Rightarrow x = v _ {0} t - \frac {v _ {0} t ^ {2}}{1 0}$
where x can be either +10 cm or -10 cm
For $x = +10 \, \mathrm{cm}$ , we have $10t - t^2 = 10$
$\Rightarrow \quad t = \left(\frac {1 0 \pm \sqrt {6 0}}{2}\right) s$
Hence we have
$t _ {1} = \left(\frac {1 0 - \sqrt {6 0}}{2}\right) \mathrm{s}$
{First Instant}
$\Rightarrow t _ {2} = \left(\frac {1 0 + \sqrt {6 0}}{2}\right) \mathrm{s}$
{Second Instant}
For $x = -10\mathrm{cm}$ , we have $10t - t^2 = -10$
$\Rightarrow t = \left(\frac {1 0 \pm \sqrt {1 4 0}}{2}\right) s$
$\Rightarrow t _ {3} = \left(\frac {1 0 + \sqrt {1 4 0}}{2}\right) \mathrm{s}$
{Third Instant}
However, we will not get fourth instant as time cannot be negative.
Time interval between the second and the third instant is $\Delta t = t_{3} - t_{2} \approx 2$ s
Q555
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
An athlete takes 2 s to reach his maximum speed of $36 \, kmh^{-1}$ . The magnitude of his average acceleration (in $ms^{-2}$ ) is .....
Correct Answer: 5
Explanation:
Initial velocity is $v_{i}=0$ and final velocity of the athlete
is $v_{f} = 36 \times \frac{5}{18} = 10 \, \mathrm{ms}^{-1}$ .
Since the particle is moving with uniform acceleration, so, we have
$v _ {a v} = \langle v \rangle = \frac {v _ {i} + v _ {f}}{2} = 5 \mathrm{ms} ^ {- 1}$
is $v_{f} = 36 \times \frac{5}{18} = 10 \, \mathrm{ms}^{-1}$ .
Since the particle is moving with uniform acceleration, so, we have
$v _ {a v} = \langle v \rangle = \frac {v _ {i} + v _ {f}}{2} = 5 \mathrm{ms} ^ {- 1}$
Q556
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The diagram shows variation of $\frac{1}{v}$ with respect to time (where v is in ms $^{-1}$ ).
Find the instantaneous acceleration $\left(\text{in } \frac{m}{s^{2}}\right)$ of body at t = 3 s.
Find the instantaneous acceleration $\left(\text{in } \frac{m}{s^{2}}\right)$ of body at t = 3 s.
Correct Answer: 3
Explanation:
3
Q557
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A ball is thrown upwards from the foot of a tower. The ball crosses the top of tower twice after an interval of 4 seconds and the ball reaches ground after 8 seconds, then the height of tower in meters is $(g = 10 \, \text{ms}^{-2})$
Correct Answer: 60
Explanation:
$\begin{array}{r l} h = u t - \frac {1}{2} g t ^ {2} \\ & \Rightarrow g t ^ {2} - 2 u t + 2 h = 0 \\ & \Rightarrow t _ {1} t _ {2} = \frac {2 h}{g} \text { and } t _ {1} + t _ {2} = \frac {2 u}{g} = T \\ & \text { Since } (t _ {2} - t _ {1}) ^ {2} = (t _ {1} + t _ {2}) ^ {2} - 4 t _ {1} t _ {2} \\ & \Rightarrow 1 6 = 6 4 - 4 \left(\frac {2 h}{g}\right) \\ & \Rightarrow h = 6 0 \mathrm{m} \end{array}$
Q558
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
An insect moves with a constant velocity v from one corner of a room to other corner which is opposite of the first corner along the largest diagonal of room. If the insect cannot fly and dimensions of room is $a \times a \times a$ , then the minimum time in which the insect can move is $\frac{a}{v}$ times the square root of a number n, then n is equal to?
Correct Answer: 5
Explanation:
$(\Delta S) _ {\min} = \left(\sqrt {a ^ {2} + \frac {a ^ {2}}{4}}\right) \times 2 = a \sqrt {5}$
$\begin{array}{r l} \Rightarrow & t _ {\min} = \frac {(\Delta s) _ {\min}}{v} = \frac {a \sqrt {5}}{v} \\ \Rightarrow & n = 5 \end{array}$
$\begin{array}{r l} \Rightarrow & t _ {\min} = \frac {(\Delta s) _ {\min}}{v} = \frac {a \sqrt {5}}{v} \\ \Rightarrow & n = 5 \end{array}$
Q559
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A body moves with constant acceleration covers 16 m and 24 m in successive intervals of 4 s and 2 s. Then its acceleration in $ms^{-2}$ is .....
Correct Answer: 4
Explanation:
4
Q560
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Figure shows the graph of velocity versus time for a particle going along x-axis. Initially at t=0, particle is at x=3 m. The position of the particle at t=2 s (in m) is ....
Correct Answer: 9
Explanation:
From graph, we have
$\begin{array}{r l} & v = t + 2 \\ \Rightarrow & x _ {2} - x _ {0} = \frac {1}{2} \times (2 + 4) \times 2 = 6 \mathrm{m} \\ \Rightarrow & x _ {2} = 9 \mathrm{m} \end{array}$
$\begin{array}{r l} & v = t + 2 \\ \Rightarrow & x _ {2} - x _ {0} = \frac {1}{2} \times (2 + 4) \times 2 = 6 \mathrm{m} \\ \Rightarrow & x _ {2} = 9 \mathrm{m} \end{array}$
Q561
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A swimmer jumps from a bridge over a canal and swims 1 km up stream. After that first km, he passes a floating cork. He continues swimming for half an hour and then turns around and swims back to the bridge. The swimmer and the cork reach the bridge at the same time. Assuming the swimmer had been swimming at a constant speed, calculate how fast does the water in the canal flow in kmh $^{-1}$ .
Correct Answer: 1
Explanation:
Let $v_{r}=u$ and $v_{sr}=v$ , then
Time taken by swimmer to go from M to O and O to B is equal to the time taken by cork to go from M to B.
$\begin{array}{r l} & {\frac {1}{2} + \frac {1 + \frac {v - u}{2}}{v + u} = \frac {1}{u}} \\ {\Rightarrow} & {\frac {1}{2} + \frac {2 + v - u}{2 (v + u)} = \frac {1}{u}} \\ {\Rightarrow} & {\frac {(v + u + 2 + v - u)}{2 (v + u)} = \frac {1}{u}} \\ {\Rightarrow} & {(2 v + 2) u = 2 (v + u)} \\ {\Rightarrow} & {2 v u + 2 u = 2 v + 2 u} \\ {\Rightarrow} & {u = 1 \mathrm{kmh} ^ {- 1}} \end{array}$
Time taken by swimmer to go from M to O and O to B is equal to the time taken by cork to go from M to B.
$\begin{array}{r l} & {\frac {1}{2} + \frac {1 + \frac {v - u}{2}}{v + u} = \frac {1}{u}} \\ {\Rightarrow} & {\frac {1}{2} + \frac {2 + v - u}{2 (v + u)} = \frac {1}{u}} \\ {\Rightarrow} & {\frac {(v + u + 2 + v - u)}{2 (v + u)} = \frac {1}{u}} \\ {\Rightarrow} & {(2 v + 2) u = 2 (v + u)} \\ {\Rightarrow} & {2 v u + 2 u = 2 v + 2 u} \\ {\Rightarrow} & {u = 1 \mathrm{kmh} ^ {- 1}} \end{array}$
Q562
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A car travelling at $60 \, kmh^{-1}$ over takes another car travelling at $42 \, kmh^{-1}$ . Assuming each car to be 5 m long. Find the time taken during the over take. (in sec)
Correct Answer: 2
Explanation:
$s_{A/B}=v_{A/B}t+\frac{1}{2}a_{A/B}t^{2}$
Q563
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is moving on a straight line with constant retardation of $1 \, ms^{-2}$ . What is the average speed of the particle on the last two meters before it stops? (in $ms^{-1}$ )
Correct Answer: 1
Explanation:
Displacement in last two seconds is
$\begin{array}{r l} & s = \Delta x = \frac {1}{2} (1) (2) ^ {2} = 2 \mathrm{m} \\ \Rightarrow & v _ {a v} = \frac {\Delta x}{\Delta t} = 1 \mathrm{ms} ^ {- 1} \end{array}$
$\begin{array}{r l} & s = \Delta x = \frac {1}{2} (1) (2) ^ {2} = 2 \mathrm{m} \\ \Rightarrow & v _ {a v} = \frac {\Delta x}{\Delta t} = 1 \mathrm{ms} ^ {- 1} \end{array}$
Q564
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A bullet going with speed $16 \, ms^{-1}$ enters a concrete wall and penetrates a distance of 0.4 m before coming to rest. Then the time taken during the retardation is $x \times 10^{-2} \, s$ . Find x.
Correct Answer: 5
Explanation:
$0 = 16^{2} - 2as$
$\left\{\because v ^ {2} = u ^ {2} + 2 a s \right\}$
$\begin{array}{r l} \Rightarrow & a = \frac {1 6 \times 1 6}{2 \times 0 . 4} = 3 2 0 \mathrm{ms} ^ {- 2} \\ \Rightarrow & t = \frac {v}{a} = \frac {1 6}{3 2 0} = 5 \times 1 0 ^ {- 2} \mathrm{s} \\ \Rightarrow & x = 5 \end{array}$
$\left\{\because v ^ {2} = u ^ {2} + 2 a s \right\}$
$\begin{array}{r l} \Rightarrow & a = \frac {1 6 \times 1 6}{2 \times 0 . 4} = 3 2 0 \mathrm{ms} ^ {- 2} \\ \Rightarrow & t = \frac {v}{a} = \frac {1 6}{3 2 0} = 5 \times 1 0 ^ {- 2} \mathrm{s} \\ \Rightarrow & x = 5 \end{array}$
Q565
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A baseball is moving at $25\mathrm{ms}^{-1}$ when it is struck by a bat and moves off in the opposite direction at $35\mathrm{ms}^{-1}$ . If the impact lasted $0.010\mathrm{s}$ , find the baseball's acceleration during the impact. (in $\mathrm{kms}^{-2}$ )
Correct Answer: 6
Explanation:
$a=\frac{v_{f}-v_{i}}{\Delta t}=\frac{35+25}{0.01}=6000\ ms^{-2}$ $\Rightarrow \quad a = 6 \, kms^{-2}$
Q566
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A point moves with uniform acceleration and its initial speed and final speed are $2 \, ms^{-1}$ and $8 \, ms^{-1}$ respectively then, the space average of velocity (in $ms^{-1}$ ) over the distance moved is .....
Correct Answer: 5.6
Explanation:
$\left[v_{av}\right]_{x}=\frac{x_{1}}{x_{2}-x_{1}}=\frac{0}{x}=5.6\ ms^{-1}$
Q567
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A man is running with a speed $8\mathrm{ms}^{-1}$ constant in magnitude and direction passes under a lantern hanging at a height $10\mathrm{m}$ above the ground. Find the velocity which the edge of the shadow of the man's head moves over the ground with if his height is $2\mathrm{m}$ .
Correct Answer: 10
Explanation:
$x_{\mathrm{man}} = vt$
$\Rightarrow \frac {O A}{A D} = \frac {B C}{B D}$
$\begin{array}{r l} \Rightarrow & \frac {1 0 \mathrm{m}}{x _ {\mathrm{man}}} = \frac {2 \mathrm{m}}{x _ {\mathrm{head}} - x _ {\mathrm{man}}} \\ \Rightarrow & x _ {\mathrm{head}} = \frac {5}{4} x _ {\mathrm{man}} = \frac {5}{4} v t \end{array}$
Velocity of shadow of head = 10 ms $^{-1}$
$\Rightarrow \frac {O A}{A D} = \frac {B C}{B D}$
$\begin{array}{r l} \Rightarrow & \frac {1 0 \mathrm{m}}{x _ {\mathrm{man}}} = \frac {2 \mathrm{m}}{x _ {\mathrm{head}} - x _ {\mathrm{man}}} \\ \Rightarrow & x _ {\mathrm{head}} = \frac {5}{4} x _ {\mathrm{man}} = \frac {5}{4} v t \end{array}$
Velocity of shadow of head = 10 ms $^{-1}$
Q568
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Two men P and Q are standing at corners A and B of square ABCD of side 8 m. They start moving along the tank with constant speed $2 \, ms^{-1}$ and $10 \, ms^{-1}$ respectively. Find the time, in second, when they will meet for the first time.
Correct Answer: 3
Explanation:
Relative displacement = relative velocity Ć time $\Rightarrow$ $8 \times 3 = (10 - 2)t$ $\Rightarrow$ t = 3 s
Q569
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle starting from rest undergoes acceleration given by $a=|t-2|$ $ms^{-2}$ where t is time in sec. Velocity of particle after 4 sec is .....
Correct Answer: 4
Explanation:
$v = \int_{0}^{4} a d t = \int_{0}^{2} a d t + \int_{2}^{4} a d t$ $\Rightarrow v = \int_{0}^{2} (2 - t) d t + \int_{2}^{4} (t - 2) d t$ $\Rightarrow v = 4 \mathrm{~ms}^{-1}$
Q570
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A ball is thrown upward from the edge of a cliff with an initial velocity of $6 \, ms^{-1}$ . How fast is it moving half second later? ( $g = 10 \, ms^{-2}$ )
Correct Answer: 1
Explanation:
$v = u + at$
$\Rightarrow v = 6 - \frac {1}{2} \times 1 0 = 1 \mathrm{ms} ^ {- 1}$
$\Rightarrow v = 6 - \frac {1}{2} \times 1 0 = 1 \mathrm{ms} ^ {- 1}$
Q571
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A car goes from 20 to $30 \, kmh^{-1}$ in 1.5 s. At the same acceleration, how long will it take the car to go from 30 to $36.7 \, kmh^{-1}$ ? (in sec)
Correct Answer: 1
Explanation:
$\begin{array}{l} {a = \frac {3 0 - 2 0}{1 . 5} = \frac {3 6 . 7 - 3 0}{\Delta t}} \\ {\Rightarrow \quad \Delta t = \frac {6 . 7 \times 1 . 5}{1 0} \approx 1 \mathrm{s}} \end{array}$
Q572
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INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The particle moves with rectilinear motion given the acceleration-displacement (a-S) curve is shown in figure. If the initial velocity is $10 \, ms^{-1}$ then velocity of particle after particle has travelled 30 m divided by 5 is equal to.
Correct Answer: 4
Explanation:
Area under curve is
$A = \frac {1}{2} \times 1 0 \times 3 0 = 1 5 0$
... (1)
Also, area under $a - x$ curve is equal to
$A = \frac {v ^ {2} - u ^ {2}}{2}$
... (2)
From (1) and (2), we get
$\begin{array}{r l} & {\frac {1}{2} \big (v ^ {2} - u ^ {2} \big) = 1 5 0} \\ {\Rightarrow} & {v ^ {2} = u ^ {2} + 3 0 0} \\ {\Rightarrow} & {v ^ {2} = (1 0) ^ {2} + 3 0 0} \\ {\Rightarrow} & {v = \sqrt {4 0 0} = 2 0 \mathrm{ms} ^ {- 1}} \end{array}$
$A = \frac {1}{2} \times 1 0 \times 3 0 = 1 5 0$
... (1)
Also, area under $a - x$ curve is equal to
$A = \frac {v ^ {2} - u ^ {2}}{2}$
... (2)
From (1) and (2), we get
$\begin{array}{r l} & {\frac {1}{2} \big (v ^ {2} - u ^ {2} \big) = 1 5 0} \\ {\Rightarrow} & {v ^ {2} = u ^ {2} + 3 0 0} \\ {\Rightarrow} & {v ^ {2} = (1 0) ^ {2} + 3 0 0} \\ {\Rightarrow} & {v = \sqrt {4 0 0} = 2 0 \mathrm{ms} ^ {- 1}} \end{array}$
Q573
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A police jeep is chasing a culprit going on a motor bike. The motor bike crosses a turning at a speed of $72 \, kmh^{-1}$ . The jeep follows it a speed of $90 \, kmh^{-1}$ crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike? (in km)
Correct Answer: 1
Explanation:
Speed of bike is
$v _ {b} = 7 2 \times \frac {5}{1 8} = 2 0 \mathrm{ms} ^ {- 1}$
Speed of jeep is
$v _ {j} = 9 0 \times \frac {5}{1 8} = 2 5 \mathrm{ms} ^ {- 1}$
Relative velocity of jeep w. r. t. bike is
$v _ {j b} = 2 5 - 2 0 = 5 \mathrm{ms} ^ {- 1}$
Distance covered by bike in 10 s is
$s _ {b} = 2 0 \times 1 0 = 2 0 0 \mathrm{m}$
Time taken by Jeep to cover 200 m with velocity $5 \, ms^{-1}$ is
$t = \frac {2 0 0}{5} = 4 0 \mathrm{s}$
Therefore distance covered by police jeep in 40 s is
$s _ {j} = 4 0 \times 2 5 = 1 0 0 0 \mathrm{m} = 1 \mathrm{km}$
$v _ {b} = 7 2 \times \frac {5}{1 8} = 2 0 \mathrm{ms} ^ {- 1}$
Speed of jeep is
$v _ {j} = 9 0 \times \frac {5}{1 8} = 2 5 \mathrm{ms} ^ {- 1}$
Relative velocity of jeep w. r. t. bike is
$v _ {j b} = 2 5 - 2 0 = 5 \mathrm{ms} ^ {- 1}$
Distance covered by bike in 10 s is
$s _ {b} = 2 0 \times 1 0 = 2 0 0 \mathrm{m}$
Time taken by Jeep to cover 200 m with velocity $5 \, ms^{-1}$ is
$t = \frac {2 0 0}{5} = 4 0 \mathrm{s}$
Therefore distance covered by police jeep in 40 s is
$s _ {j} = 4 0 \times 2 5 = 1 0 0 0 \mathrm{m} = 1 \mathrm{km}$
Q574
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of $1 \, ms^{-2}$ and projection velocity in the vertical direction is $9.8 \, ms^{-1}$ . How far behind the boy will the ball fall on the car? (in m)
Correct Answer: 2
Explanation:
Time when velocity of ball is zero
$0 = 9. 8 \times g t$
$\Rightarrow t = \frac {9 . 8}{9 . 8} = 1 \mathrm{s}$
So, total time taken by ball to come back is 2 s.
Distance travelled by trolley in 2 s is
$s = \frac {1}{2} a t ^ {2} = \frac {1}{2} \times 1 \times 2 ^ {2} = 2 \mathrm{m}$
So, ball will fall 2 m behind the boy.
$0 = 9. 8 \times g t$
$\Rightarrow t = \frac {9 . 8}{9 . 8} = 1 \mathrm{s}$
So, total time taken by ball to come back is 2 s.
Distance travelled by trolley in 2 s is
$s = \frac {1}{2} a t ^ {2} = \frac {1}{2} \times 1 \times 2 ^ {2} = 2 \mathrm{m}$
So, ball will fall 2 m behind the boy.
Q575
Advance
INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The speed of a motor launch with respect to the water is $v = 5 \, ms^{-1}$ , the speed of stream $u = 3 \, ms^{-1}$ . When the launch began travelled 3.6 km up stream, turned about and caught up with the float. How long is it before the launch reaches the float again? (Find answer in hour).
Correct Answer: 1
Explanation:
$t=\frac{2\ell}{v-u}=\frac{2\times3600}{2}$
$\Rightarrow t = 3 6 0 0 \mathrm{s}$
$\Rightarrow t = 1 \mathrm{h}$
$\Rightarrow t = 3 6 0 0 \mathrm{s}$
$\Rightarrow t = 1 \mathrm{h}$