Question

A vehicle starts moving in a straight line with an acceleration, $a=4\, m / s ^{2}$, with initial velocity equal to zero. After accelerating for time $t_{1}$, the vehicle moves uniformly and for time $t_{2}$, the vehicle finally decelerates for time $t_{1}$ eventually coming to a stop. The total time taken during the motion is $10\, s$ and the average velocity during the motion is $5.1\, m / s$. The time taken by the vehicle during acceleration is

• A. 2 s
• B. 2.5 s
• C. 1.5 s
• D. 1.8 s

Answer 1

Answer: (C)

For first part of journey, $x=0, a=4\, ms ^{-2}$, time $=t_{1}$.
Velocity attained by particle at end of time $t_{1}$ is $v_{1}=u +a t_{1}=4 t_{1}$
$\therefore $ Displacement in first $t_{1}$ seconds is
$s_{1}=\frac{1}{2} a t_{1}^{2}=\frac{1}{2} \times 4 \times t_{1}^{2}=2 t_{1}^{2}$
For second part of journey, initial velocity, $v_{1}=4 t_{1}$, time $=t_{2}$ and acceleration $=0$
$\therefore $ Displacement in next $t_{2}$ seconds is $s_{2}=v_{1} t_{2}=4 t_{1} t_{2}$
For third part of journey,
Initial velocity, $v_{1}=4 t_{1}$
Final velocity $=0$
Time interval $=t_{1}$
So, acceleration $=\frac{0-4 t_{1}}{t_{1}}=-4\, ms ^{-2}$
$\therefore $ Displacement in third part is
$s_{3} =\frac{-16 t_{1}^{2}}{-8}=2 t_{1}^{2} \left(\because v_{2}^{2}-v_{1}^{2}=2 a s_{3}\right)$
Now, $v_{\text {avg }}=\frac{\text { Total displacement }}{\text { Total time }}$
$5.1=\frac{s_{1}+s_{2}+s_{3}}{t_{1}+t_{2}+t_{1}}$
$\left[\therefore \text { third time interval }=t_{1}\right]$
$5.1=\frac{2 t_{1}^{2}+4 t_{1} t_{2}+2 t_{1}^{2}}{10}$
$\Rightarrow 51=4 t_{1}^{2}+4 t_{1}\left(10-2 t_{1}\right)$
$\Rightarrow 51=4 t_{1}^{2}+40 t_{1}-8 t_{1}^{2}$
$\Rightarrow 4 t_{1}^{2}-40 t_{1}+51=0$
$\Rightarrow 4 t_{1}^{2}-34 t_{1}-6 t_{1}+51=0$
$\Rightarrow \left(2 t_{1}-3\right)\left(2 t_{1}-17\right)=0$
$\Rightarrow t_{1}=\frac{3}{2}$
or $t_{1}=\frac{17}{2} \Rightarrow t_{1}=1.5 s$
(As $t_{1}=8.5 s$ is not possible. It exceeds total time.)