Motion in a Straight Line

Given, initial velocity of truck,

$ \begin{aligned} u=54 \mathrm{~km} / \mathrm{h} & \\ & =54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s} \end{aligned} $

Final velocity, $v=0$

Deceleration, $a=-10 \mathrm{~m} / \mathrm{s}^2$

As we know that,

$ \begin{aligned} v^2 & =u^2+2 a s \\ \Rightarrow \quad(0)^2 & =(15)^2+2 \times(-10) \mathrm{s} \\ \Rightarrow \quad 20 \mathrm{~s} & =(15)^2 \\ s & =\frac{15 \times 15}{20}=11.25 \mathrm{~m} \end{aligned} $

When ball is at maximum height, its velocity is zero. From first equation of motion,

$ \begin{aligned} & & v & =u-g t \Rightarrow 0=u-10 \times 5 \\ \Rightarrow & & u & =50 \mathrm{~m} / \mathrm{s} \end{aligned} $

Distance travelled in $n$th second,

$ s_n=u+\frac{a}{2}(2 n-1) $

Distance travelled in 2nd second is given as

$ s_2=50-\frac{10}{2}(4-1)=50-15=35 \mathrm{~m} $

Distance travelled in 7th second is equal to distance travelled in 2nd second from highest point to vertically downward direction.

$ \begin{aligned} s_7 & =s_2^{\prime}=0+\frac{10}{2}(2 \times 2-1)=15 \mathrm{~m} \\ \therefore \quad & \frac{s_2}{s_7}=\frac{s_2}{s_2^{\prime}}=\frac{35}{15}=\frac{7}{3} \text { or } 7: 23 \end{aligned} $

The given situation of entire motion of ball is shown in the following figure.

Initial speed of ball, $u=20 \mathrm{~m} / \mathrm{s}$

Total time taken by the ball to reach at highest point $P$ is $t_1$, then from equation of vertical motion

$ \begin{aligned} & v = u-g t \\ \Rightarrow& 0=20-10 t_1 \Rightarrow t_1=2 \mathrm{~s} \end{aligned} $

Maximum height attained by the ball is $h$, then from equation

$ \begin{aligned} h & =u t-\frac{1}{2} g t^2 \\ & =20 \times 2-\frac{1}{2} \times 10 \times 2^2 \\ & =40-20=20 \mathrm{~m} \end{aligned} $

If $t_2$ be the time taken by the ball to reach at $B$ from point $P$, then

$ \begin{aligned} h-5 & =0 \times t_2+\frac{1}{2} g t_2^2 \\ \Rightarrow \quad 20-5 & =0+\frac{1}{2} \times 10 t_2^2 \\ 15 & =5 t_2^2 \Rightarrow t_2=\sqrt{3} \mathrm{~s} \end{aligned} $

∴ Time taken by the ball during entire trip,

$ t=t_1+t_2=2+\sqrt{3} \mathrm{~s} $

The given situation of entire motion of motor bike is shown in the figure.

If time taken by the motorbike to reach from point $A$ to $B$ is $t_{A B}$, then for $A$ to $B$,

$ v_A=0, v_B=10 \mathrm{~ms}^{-1}, a=0.5 \mathrm{~ms}^{-2} $

∴ From equation, $v=u+$ at

$ \begin{aligned} & & v_B & =v_A+a t_{A B} \\ \Rightarrow & & 10 & =0+0.5 t_{A B} \\ \Rightarrow & & t_{A B} & =\frac{10}{0.5}=20 \mathrm{~s} \end{aligned} $

Since, motorbike moves from point $B$ to $C$ with constant velocity.

Hence, time taken to travel distance $B C$ is given as

$ \begin{aligned} t_{B C} & =\frac{B C}{v_B}=\frac{10 \mathrm{~km}}{10 \mathrm{~ms}^{-1}} \\ & =\frac{10000 \mathrm{~m}}{10 \mathrm{~ms}^{-1}}=1000 \mathrm{~s} \end{aligned} $

For motion of motorbike from point $C$ to $D$, time $=t_{C D}, v_B=10 \mathrm{~ms}^{-1}$,

$ v_f=0, a^{\prime}=0.2 \mathrm{~ms}^{-2} $

From equation,

$ \begin{aligned} v & =u-a t, v_f=v_B-a^{\prime} t_{C D} \\ 0 & =10-0.2 \times t_{C D} \\ \Rightarrow \quad t_{C D} & =50 \mathrm{~s} \\ \therefore \text { Total time } & =t_{A B}+t_{B C}+t_{C D} \\ & =20 \mathrm{~s}+1000 \mathrm{~s}+50 \mathrm{~s} \\ & =1070 \mathrm{~s} \end{aligned} $

The position of the particle is given as

$ y(t)=10 t e^{-2 t} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Velocity of the particle is given as

$ \begin{aligned} v & =\frac{d}{d t} y(t)=\frac{d}{d t}\left(10 t e^{-2 t}\right) \\ & =10\left[t e^{-2 t}(-2)+e^{-2 t} \cdot 1\right] \\ & =10\left[-2 t e^{-2 t}+e^{-2 t}\right]=10 e^{-2 t}(1-2 t) \end{aligned} $

When body stops, then

$ \begin{aligned} v & =0 \Rightarrow e^{-2 t}(1-2 t)=0 \\ 1-2 t & =0 \\ t & =\frac{1}{2} \mathrm{~s} \end{aligned} $

∴ From Eq. (i), displacement of the particle from origin is given as

$ y\left(\frac{1}{2}\right)=10 \times \frac{1}{2} \times e^{-2 \times \frac{1}{2}}=5 e^{-1}=\frac{5}{e} \mathrm{~m} $

The motion of two car $A$ and $B$ is shown in the figure.

Relative velocity of car $A$ w.r.t. car $B$ is

$ v_{A B}=v_A-v_B=15-10=5 \mathrm{~m} / \mathrm{s} $

Distance to be covered for overtake is given as

$ s=4 \mathrm{~m}+4 \mathrm{~m}=8 \mathrm{~m} $

Time taken by cars to cross

$ \begin{aligned} & =\text { Distance/Speed } \\ & =\frac{s}{V_{A B}}=\frac{8}{5}=1.6 \mathrm{~s} \end{aligned} $

An automobile traveling at 40 km/h can be stopped within a distance of 40 meters by using brakes. If the same automobile is traveling at 80 km/h, we need to determine the minimum stopping distance in meters, assuming no skidding occurs.

First Case

• Initial speed, $ u_1 = 40 \, \text{km/h} $


• Final speed, $ v_1 = 0 \, \text{m/s} $


• Stopping distance, $ s_1 = 40 \, \text{m} $

Using the equation:

$ v^2 - u^2 = 2as $

For the first case:

$ 0^2 - 40^2 = 2a \times 40 $

$ -1600 = 80a $

$ a = -20 \, \text{m/s}^2 $

Second Case

• Initial speed, $ u_2 = 80 \, \text{km/h} $


• Final speed, $ v_2 = 0 \, \text{m/s} $

Similarly, using the same equation:

$ 0^2 - 80^2 = 2a s_2 $

$ -6400 = 2a s_2 $

Since $ a = -20 \, \text{m/s}^2 $, divide both sides of the equation for the second case by the first case:

$ \frac{s_2}{40} = \frac{80^2}{40^2} $

$ s_2 = \frac{80 \times 80}{40} $

$ s_2 = 160 \, \text{m} $

Thus, the minimum stopping distance when the automobile is traveling at 80 km/h is 160 meters.

Given the velocity function: $v = \alpha \sqrt{x}$

We know that velocity $v$ is the rate of change of displacement with respect to time:

$\therefore \frac{dx}{dt} = \alpha \sqrt{x}$

Rearranging and separating variables, we get:

$\Rightarrow \frac{dx}{\sqrt{x}} = \alpha \, dt$

To solve this, we integrate both sides:

$\int\limits_{0}^{x} \frac{dx}{\sqrt{x}} = \alpha \int\limits_{0}^{t} dt$

This gives us:

$\Rightarrow \left[ \frac{2\sqrt{x}}{1} \right]_{0}^{x} = \alpha \left[t\right]_{0}^{t}$

Simplifying, we obtain:

$\Rightarrow 2\sqrt{x} = \alpha t $

Squaring both sides:

$\Rightarrow x = \frac{\alpha^2}{4} t^2$

Thus, the displacement $x$ varies with time $t$ as:

$x \propto t^2$

The relationship between time $ t $ and distance $ x $ is defined as $ t = ax^2 + bx $, where $ a $ and $ b $ are constants.

To find the acceleration, follow these steps:

First, differentiate the equation with respect to time $ t $:

$ \frac{d}{dt}(t) = a \frac{d}{dt}(x^2) + b \frac{dx}{dt} $

This simplifies to:

$ 1 = a \cdot 2x \frac{dx}{dt} + b \frac{dx}{dt} $

Given $ \frac{dx}{dt} = v $ (where $ v $ is the velocity), we can write:

$ 1 = 2axv + bv = v(2ax + b) $

This simplifies to:

$ 2ax + b = \frac{1}{v} $

Next, differentiate the expression $ 2ax + b = \frac{1}{v} $ again with respect to $ t $:

$ 2a \frac{dx}{dt} + 0 = -\frac{1}{v^2} \frac{dv}{dt} $

Using $ \frac{dx}{dt} = v $, we get:

$ 2av = -\frac{1}{v^2} \frac{dv}{dt} $

Solving for $ \frac{dv}{dt} $:

$ \frac{dv}{dt} = -2a v^3 $

Since acceleration $ f = \frac{dv}{dt} $, we have:

$ f = -2av^3 $

Given the initial speeds of two identical cars as $u$ and $4u$, and considering that both cars eventually stop (final speed $v = 0$), we note that both cars decelerate with the same acceleration $-a$.

Using the kinematic equation:

$ v^2 = u^2 - 2as $

Since $v = 0$,

$ 0 = u^2 - 2as $

Hence,

$ u^2 = 2as $

For the first car with speed $u$,

$ u^2 = 2a s_1 \quad \text{...(i)} $

For the second car with speed $4u$,

$ (4u)^2 = 2a s_2 \quad \text{...(ii)} $

Dividing equation (i) by equation (ii),

$ \frac{u^2}{(4u)^2} = \frac{2a s_1}{2a s_2} $

$ \frac{u^2}{16u^2} = \frac{s_1}{s_2} $

$ \frac{1}{16} = \frac{s_1}{s_2} $

Thus, the ratio of the stopping distances of the two cars is $\frac{1}{16}$.