Question

A vehicle starts moving in a straight line with an acceleration, $a=4m/s^2$, 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 $10s$ and the average velocity during the motion is $5.1m/s$. 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, $x=0,a=4m{s}^{-2}$, time $=t_1$.
Velocity attained by particle at end of time $t_1$ is $v_1=u+a{t}_1=4t_1$
$\therefore$ Displacement in first $t_1$ seconds is
$s_1=\frac{1}{2}a{t}_1^2=\frac{1}{2}\times 4\times t_1^2=2t_1^2$
For second part of journey, initial velocity, $v_1=4t_1$, time $=t_2$ and acceleration $=0$
$\therefore$ Displacement in next $t_2$ seconds is $s_2=v_1{t}_2=4t_1{t}_2$
For third part of journey,
Initial velocity, $v_1=4t_1$
Final velocity $=0$
Time interval $=t_1$
So, acceleration $=\frac{0-4t_1}{t_1}=-4m{s}^{-2}$
$\therefore$ Displacement in third part is
$s_3=\frac{-16t_1^2}{-8}=2t_1^2\left(\because v_2^2-v_1^2=2a{s}_3\right)$
Now, $v_{\text{avg }}=\frac{\text{ Total displacement }}{\text{ Total time }}$
$5.1=\frac{s_1+s_2+s_3}{t_1+t_2+t_1}$
$\left[\therefore \text{ third time interval }=t_1\right]$
$5.1=\frac{2t_1^2+4t_1{t}_2+2t_1^2}{10}$
$\Rightarrow 51=4t_1^2+4t_1\left(10-2t_1\right)$
$\Rightarrow 51=4t_1^2+40t_1-8t_1^2$
$\Rightarrow 4t_1^2-40t_1+51=0$
$\Rightarrow 4t_1^2-34t_1-6t_1+51=0$
$\Rightarrow \left(2t_1-3\right)\left(2t_1-17\right)=0$
$\Rightarrow t_1=\frac{3}{2}$
or $t_1=\frac{17}{2}\Rightarrow t_1=1.5s$
(As $t_1=8.5s$ is not possible. It exceeds total time.)