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 $20m/s$, then find the value of $t$.

• A. $6s$
  
• B. $5s$
  
• C. $9s$
  
• D. $4s$
  

Answer 1

The correct option is B $5s$

average velocity is $25m/s$ so distance will be $25\times 20=500m$

now finding area of the graph between velocity and time graph we get

area of two triangles + area of rectangle.

$2.\frac{1}{2}t.5.t+(25-2t).5t=500=>t^2-25t+100$

$=>t=5$

so option B is answer.