Question

The velocity - time graph of the particle moving along a straight line is shown. The rate of acceleration and deceleration is constant and it is equal to $5m{s}^{-2}$. If the average velocity during the motion is $20m{s}^{-1}$, then the value of $t$ is

• A. $3sec$
• B. $5sec$
• C. $10sec$
• D. $12sec$

Answer 1

The correct option is B $5sec$

$\frac{v_0}{t}=\frac{v_0}{25-t-t_1}=5$...(1)

$t=25-t-t_1$

$2t=25-t_1$

$t_1=25-2t$

$s_1=\frac{1}{2}{v}_{0}t$

$s_2=v_0{t}_1=v_0(25-2t)$

$s_3=\frac{1}{2}\times t\times v_0$

$\frac{\left(\frac{v_{0}t}{2}\right)\times 2+v_0(25-2t)}{25}=20\frac{v_{0}t+25v_0-2v_{0}t}{25}=20\frac{25v_0-v_{0}t}{25}=20v_0=5t25\times 5t-5t^2=25\times 20t^2-25t+100=0t^2-20t-5t+100=0t(t-20)-5(t-20)=0$

$t=5s$ or $t=20s$

$t=5s$