Question

Car accelerates from rest at a constant rate $2am/s^2$ for sometime and attains a velocity of $20m/s$. Afterwards it decelerates with a constant rate $am/s^2$ and comes to a halt. If the total time taken is $10s$, the distance traveled by the car is :

• A. $100m$
  
• B. $200m$
  
• C. $300m$
  
• D. $400m$
  

Answer 1

The correct option is A $100m$

If $t_1$ is the time for acceleration and $t_2$ is for deceleration, $t_1+t_2=10$ ...(1)

Using formula, $v=u+at$,

for acceleration, $20=0+2a{t}_1$ or $t_1=10/a$

for deceleration, $0=20-a{t}_2$ or $t_2=20/a$

From (1), $10/a+20/a=10$ or $a=3$

Now, distance traveled in time $t_1$ is $s_1=0\left(t_1\right)+\frac{1}{2}\left(2a\right)t_1^2=a{t}_1^2=a(10/a{)}^2=100/a=100/3$

and distance traveled in time $t_2$ is given by, $0^2-{20}^2=2(-a)s_2$ (using $v^2-u^2=2as$)

or $s_2=200/a=200/3$

Thus, total distance traveled by the car $s_1+s_2=100/3+200/3=100m$