Explanation:
To find the acceleration of the particle, we first need to differentiate the velocity function with respect to time. The velocity function given is
$ v = 4\sqrt{x} $
However, this function gives the velocity as a function of position $x$, not as a function of time $t$. Since acceleration is the rate of change of velocity with respect to time, we'll need to use the chain rule to differentiate $v$ with respect to $t$.
The chain rule in this context can be stated as follows:
$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} $
Now, because $\frac{dx}{dt}$ is the velocity $v$ itself and $\frac{dv}{dx}$ is the derivative of the velocity with respect to $x$, we first find $\frac{dv}{dx}$:
$ v = 4\sqrt{x} = 4x^{\frac{1}{2}} $
Differentiating with respect to $x$, we get:
$ \frac{dv}{dx} = 4 \cdot \frac{1}{2} x^{-\frac{1}{2}} = 2x^{-\frac{1}{2}} = \frac{2}{\sqrt{x}} $
Now, because $v = 4\sqrt{x}$, we can rewrite $\sqrt{x}$ as $\frac{v}{4}$. Using this to replace $\sqrt{x}$ in our expression for $\frac{dv}{dx}$, we get:
$ \frac{dv}{dx} = \frac{2}{\sqrt{x}} = \frac{2}{\frac{v}{4}} = \frac{8}{v} $
Now, using the chain rule:
$ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{8}{v} \cdot v $
Simplifying this, the velocity terms cancel out, leaving us with:
$ a = 8 \text{ ms}^{-2} $
Thus, the acceleration of the particle is $8 \text{ ms}^{-2}$.
The velocity of the particle when its acceleration becomes zero is _________ $\mathrm{m} / \mathrm{s}$.
Explanation:
- Position (x): The particle's location on the x-axis at a given time.
- Velocity (v): The rate of change of position with respect to time
( $v = \frac{dx}{dt}$ ).
- Acceleration (a): The rate of change of velocity with respect to time
( $a = \frac{dv}{dt}$ ).
The particle's position is given by:
$x(t) = -3t^3 + 18t^2 + 16t$
We need to find the velocity when the acceleration is zero.
- Find the Velocity (v) and Acceleration (a) Functions:
- Velocity:
$v(t) = \frac{dx}{dt} = -9t^2 + 36t + 16$
- Acceleration:
$a(t) = \frac{dv}{dt} = -18t + 36$
- Find the Time (t) When Acceleration is Zero:
$a(t) = 0$
$-18t + 36 = 0$
$t = 2 \text{ seconds}$
- Calculate the Velocity at t = 2 seconds:
$v(2) = -9(2)^2 + 36(2) + 16$
$v(2) = 52 \text{ m/s}$
Answer:The velocity of the particle when its acceleration becomes zero is 52 m/s.
The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t+1) \mathrm{s}$ are $125 \mathrm{~m}$ and $50 \mathrm{~m} / \mathrm{s}$, respectively. The distance travelled by the particle in $(\mathrm{t}+2)^{\mathrm{th}} \mathrm{s}$ is _________ m.
Explanation:
The displacement and the increase in the velocity of a moving particle from time $t$ to $(t + 1) \mathrm{s}$ are $125 \mathrm{~m}$ and $50 \mathrm{~m} / \mathrm{s}$, respectively. The distance traveled by the particle in $(\mathrm{t} + 2)^{\mathrm{th}} \mathrm{s}$ is calculated as follows:
Given that the acceleration is constant, we start with:
$ v = u + at $
When the velocity has increased by $50 \mathrm{~m}/\mathrm{s}$, the equation becomes:
$ u + 50 = u + a \quad \Rightarrow \quad a = 50 \mathrm{~m}/\mathrm{s}^2 $
Next, we consider the displacement:
$ 125 = u t + \frac{1}{2} a t^2 $
Since this is given over a unit time interval (from $t$ to $(t + 1)$), we use:
$ 125 = u + \frac{a}{2} $
Substituting $a = 50 \mathrm{~m}/\mathrm{s}^2$:
$ 125 = u + \frac{50}{2} \quad \Rightarrow \quad 125 = u + 25 \quad \Rightarrow \quad u = 100 \mathrm{~m}/\mathrm{s} $
To find the distance traveled by the particle in $(t + 2)^\text{th}$ second, we use:
$ S_n = u + \frac{a}{2} [2n - 1] $
For $n = t + 2$ (i.e., the (t+2)th second):
$ S_{(t+2)} = u + \frac{a}{2} [2(t+2) - 1] $
With $u = 100$ and $a = 50$:
$ S_{(t+2)} = 100 + \frac{50}{2} [2(t+2) - 1] = 100 + 25 \times [2(t+2) - 1] = 100 + 25 \times (2t + 4 - 1) = 100 + 25 \times (2t + 3) = 100 + 25 \times 5 = 100 + 125 = 225 \mathrm{~m} $
A body falling under gravity covers two points $A$ and $B$ separated by $80 \mathrm{~m}$ in $2 \mathrm{~s}$. The distance of upper point A from the starting point is _________ $\mathrm{m}$ (use $\mathrm{g}=10 \mathrm{~ms}^{-2}$).
Explanation:
To find the distance of the upper point A from the starting point, we need to first understand the motion of a freely falling body under the influence of gravity. The body falling under gravity is an example of uniformly accelerated motion with the acceleration equal to the acceleration due to gravity, which is given as $ g = 10 \textrm{m/s}^2 $.
We will consider point A where the body was at time $ t_1 $ and point B where the body was at time $ t_2 = t_1 + 2 \textrm{s} $. The displacement in these 2 seconds is given to be 80 meters.
For an object under constant acceleration, the displacement $ s $ can be found using the equation:
$ s = ut + \frac{1}{2}at^2 $
where
$ u $ is the initial velocity,
$ t $ is the time,
$ a $ is the acceleration,
$ s $ is the displacement.
Since the body falls under gravity, its initial velocity at the starting point is 0 ($ u = 0 $), thus the equation simplifies to:
$ s = \frac{1}{2}gt^2 $
Let's denote the distance of point A from the starting point as $ s_A $ and the distance of point B as $ s_B $. We know the body covers $ 80 $ meters in $ 2 $ seconds from A to B, so we can write:
$ s_B - s_A = 80 \textrm{m} $
We can calculate the distance covered till point B (in time $ t_2 = t_1 + 2 $) as:
$ s_B = \frac{1}{2}g(t_1 + 2)^2 $
And the distance covered till point A (in time $ t_1 $) as:
$ s_A = \frac{1}{2}gt_1^2 $
Now, substituting $ s_B $ and $ s_A $ in our difference equation:
$ \frac{1}{2}g(t_1 + 2)^2 - \frac{1}{2}gt_1^2 = 80 $
$ g \left( \frac{1}{2}(t_1 + 2)^2 - \frac{1}{2}t_1^2 \right) = 80 $
$ 5((t_1 + 2)^2 - t_1^2) = 80 $
$ (t_1 + 2)^2 - t_1^2 = 16 $
$ t_1^2 + 4t_1 + 4 - t_1^2 = 16 $
$ 4t_1 + 4 = 16 $
$ 4t_1 = 16 - 4 $
$ 4t_1 = 12 $
$ t_1 = 3 \textrm{s} $
Therefore, time $ t_1 $ at point A is $ 3 $ seconds. To find $ s_A $, we can use the above simplified motion equation:
$ s_A = \frac{1}{2}gt_1^2 $
$ s_A = \frac{1}{2} \times 10 \times 3^2 $
$ s_A = 5 \times 9 $
$ s_A = 45 \textrm{m} $
So, the distance of the upper point A from the starting point is $ \mathbf{45 \textrm{m}} $.
Alternate Method :
From $\mathrm{A} \rightarrow \mathrm{B}$
$ \begin{aligned} & -80=-\mathrm{v}_1 \mathrm{t}-\frac{1}{2} \times 10 \mathrm{t}^2 \\\\ & -80=-2 \mathrm{v}_1-\frac{1}{2} \times 10 \times 2^2 \\\\ & -80=-2 \mathrm{v}_1-20 \\\\ & -60=-2 \mathrm{v}_1 \\\\ & \mathrm{v}_1=30 \mathrm{~m} / \mathrm{s} \end{aligned} $
From $\mathrm{O}$ to $\mathrm{A}$
$ \mathrm{v}^2=\mathrm{u}^2+2 \mathrm{gS} $
$ \Rightarrow $ $ 30^2=0+2 \times(-10)(-\mathrm{S}) $
$ \Rightarrow $ $ 900=20 \mathrm{~S} $
$ \Rightarrow $ $ \mathrm{S}=45 \mathrm{~m} $
For a train engine moving with speed of $20 \mathrm{~ms}^{-1}$, the driver must apply brakes at a distance of 500 $\mathrm{m}$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt{x} \mathrm{~ms}^{-1}$. The value of $x$ is ____________.
(Assuming same retardation is produced by brakes)
Explanation:
$ \begin{aligned} & v^2=u^2+2 a s \\\\ & (0)^2=u^2+2 a s \\\\ & u^2=-2 a s \\\\ & S=\frac{u^2}{2 a}-\frac{(20)^2}{2 \times a}=500 \\\\ & \text { acceleration of the train, } a=-\frac{400}{1000}=-0.4 \mathrm{~m} / \mathrm{sec} \end{aligned} $
Now, if the brakes are applied at $S=250 \mathrm{~m}$ i.e. half of the distance
$ \begin{aligned} & v^2=u^2+2 a s \\\\ & v^2=(20)^2+2(-0.4) \times 250 \\\\ & v^2=400-2 \times \frac{4}{10} \times 250 \\\\ & v^2=200 \\\\ & v=\sqrt{200} \\\\ & \text { Given } \Rightarrow v=\sqrt{x} \\\\ & \therefore x=200 \end{aligned} $
A horse rider covers half the distance with $5 \mathrm{~m} / \mathrm{s}$ speed. The remaining part of the distance was travelled with speed $10 \mathrm{~m} / \mathrm{s}$ for half the time and with speed $15 \mathrm{~m} / \mathrm{s}$ for other half of the time. The mean speed of the rider averaged over the whole time of motion is $\frac{x}{7} \mathrm{~m} / \mathrm{s}$. The value of $x$ is ___________.
Explanation:
$ \Rightarrow {t_1} = {{{S \over 2}} \over 5}$ ........ (1)
Also, ${S \over 2} = {{10{t_2}} \over 2} + {{15{t_2}} \over 2}$
$ \Rightarrow {t_2} = {S \over {25}}$ ....... (2)
$\Rightarrow$ Mean speed $ = {S \over {{t_1} + {t_2}}}$
$ = {S \over {{S \over {10}} + {S \over {25}}}} = {{250} \over {35}}$ m/s $ = {{50} \over 7}$ m/s
A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms$^{-2}$.
(Given g = 10 ms$^{-2}$)
Explanation:
$u=\sqrt{2 \times g H}=\sqrt{2 \times 10 \times 9.8}=\underset{\text { (Downwards) }}{14 \mathrm{~m} / \mathrm{sec}}$
The speed of ball just after collision is
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 5}=\underset{\text { (Upwards) }}{10 \mathrm{~m} / \mathrm{sec}}$
So, $\vec{a}=\frac{\Delta \vec{v}}{\Delta t}$
$=\frac{10+14}{0.2}=120 \mathrm{~m} / \mathrm{s}^{2}$
A ball is thrown vertically upwards with a velocity of $19.6 \mathrm{~ms}^{-1}$ from the top of a tower. The ball strikes the ground after $6 \mathrm{~s}$. The height from the ground up to which the ball can rise will be $\left(\frac{k}{5}\right) \mathrm{m}$. The value of $\mathrm{k}$ is __________. (use $\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}$)
Explanation:
v = 19.6 m/s
t = 6s
Time taken in upward motion above tower = 2s
$\Rightarrow$ Time taken from top most point to ground = 4s
$ \Rightarrow \sqrt {{{2h} \over g}} = 4$
$h = {{16 \times 9.8} \over 2} = 8 \times 9.8$
$ \Rightarrow k = 8 \times 9.8 \times 5 = 392$
A particle is moving in a straight line such that its velocity is increasing at 5 ms$-$1 per meter. The acceleration of the particle is _____________ ms$-$2 at a point where its velocity is 20 ms$-$1.
Explanation:
${{dv} \over {dx}} = 5$ ms$-$1/m
Acceleration of particle
when $v = 20$ m/s
$a = v{{dv} \over {dx}} = 20(5)$ m/s2 = 100 m/s2
A car is moving with speed of $150 \mathrm{~km} / \mathrm{h}$ and after applying the break it will move $27 \mathrm{~m}$ before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ___________ m distance.
Explanation:
${F_R}\,d = {1 \over 2}m{v^2}$
${{{d_2}} \over {{d_1}}} = {\left( {{{{v_2}} \over {{v_1}}}} \right)^2} = {\left( {{1 \over 3}} \right)^2}$
${d_2} = {d_1} \times {1 \over 9} = 3m$
A car covers AB distance with first one-third at velocity v1 ms$-$1, second one-third at v2 ms$-$1 and last one-third at v3 ms$-$1. If v3 = 3v1, v2 = 2v1 and v1 = 11 ms$-$1 then the average velocity of the car is _____________ ms$-$1.
Explanation:
${v_{mean}} = {{3{v_1}{v_2}{v_3}} \over {{v_1}{v_2} + {v_2}{v_3} + {v_3}{v_1}}}$
$ = {{3 \times 11 \times 22 \times 33} \over {11 \times 22 + 22 \times 33 + 33 \times 11}}$
$ = 18$ m/sec
A ball is projected vertically upward with an initial velocity of 50 ms$-$1 at t = 0s. At t = 2s, another ball is projected vertically upward with same velocity. At t = __________ s, second ball will meet the first ball (g = 10 ms$-$2).
Explanation:
At t = 2 s, v1 = 50 $-$ 2 $\times$ 10 = 30 m/s
v2 = v2
$\therefore$ arel = g $-$ g = 0
$S = {{{u^2} - {v^2}} \over {2g}} = {{{{50}^2} - {{30}^2}} \over {2 \times 10}} = {{1600} \over {20}} = 80$ m
$\therefore$ vrel = 50 $-$ 30 = 20 m/s
$\therefore$ $\Delta t = {{80} \over {20}} = 4\,s$
$\therefore$ required time t = 2 + 4 = 6 s
A ball of mass 0.5 kg is dropped from the height of 10 m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is ________ m. [Use g = 10 m/s2]
Explanation:
Let at h height its velocity becomes 10 m/s (as given in question)
$ \Rightarrow {v^2} - {u^2} = 2( - 10) \times ( - h)$
$ \Rightarrow {v^2} = 20\,h$
$ \Rightarrow 100 = 20\,h$
$ \Rightarrow h = 5\,m$
$h' = 10 - 5 = 5\,m$
From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____________ s.
Explanation:
Based on the situation
$h = - u{t_1} + {1 \over 2}gt_1^2$ $\to$ throwing up ....... (i)
$h = u{t_2} + {1 \over 2}gt_2^2$ $\to$ throwing up ....... (ii)
$h = {1 \over 2}g{t^2}$ $\to$ dropping .......... (iii)
and $0 = u({t_1} - {t_2}) - {1 \over 2}g{({t_1} - {t_2})^2}$ ....... (iv)
solving above equations
$t = \sqrt {{t_1}{t_2}} $
$ \Rightarrow t = \sqrt {6 \times 1.5} = 3\,s$
Explanation:
v2 = ${{20} \over {10}}$x + 20
v2 = 2x + 20
2v${{dv} \over {dx}}$ = 2
$\therefore$ a = v${{dv} \over {dx}}$ = 1
Explanation:
${{dV} \over {dx}} = {1 \over {2\sqrt {5000 + 24x} }} \times 24 = {{12} \over {\sqrt {5000 + 24x} }}$
Now, $a = V{{dV} \over {dx}}$
$ = \sqrt {5000 + 24x} \times {{12} \over {\sqrt {5000 + 24x} }}$
a = 12 m/s2
Explanation:
When both balls will collide
y1 = y2
$35t - {1 \over 2} \times 10 \times {t^2} = 35(t - 3) - {1 \over 2} \times 10 \times {(t - 3)^2}$
$35t - {1 \over 2} \times 10 \times {t^2} = 35t - 105 - {1 \over 2} \times 10 \times {t^2} - {1 \over 2} \times 10 \times {3^2} + {1 \over 2} \times 10 \times 6t$
0 = 150 $-$ 30 t
t = 5 sec
$\therefore$ Height at which both balls will collied
$h = 35t - {1 \over 2} \times 10 \times {t^2}$
$ = 35 \times 5 - {1 \over 2} \times 10 \times {5^2}$
h = 50 m
Explanation:
= ${1 \over 2} \times 8 \times 5$ = 20 m
x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is __________
Explanation:
$ \Rightarrow $ 2xv = 2at + 2b .....(2)
$ \Rightarrow $ v = ${{at + b} \over x}$
differentiating (2) w.r.t. time
$ \Rightarrow $ xa' + v2 = a
Here a' is acceleration.
$ \Rightarrow $ a'x = a - v2 = a - ${\left[ {{{at + b} \over x}} \right]^2}$
$ \Rightarrow $ a'x = ${{a{x^2} - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$
$ \Rightarrow $ a' = ${{a\left( {a{t^2} + 2bt + c} \right) - {{\left( {at + b} \right)}^2}} \over {{x^2}}}$
$ \Rightarrow $ a' = ${{ac - {b^2}} \over {{x^3}}}$
$ \therefore $ a' $ \propto $ ${1 \over {{x^3}}}$ $ \propto $ x-3
$ \therefore $ n = 3