Question

A car accelerates with an acceleration of $2{\text{m/s}}^2$ for $20\text{s}$ from rest on a straight path. Then it decelerates with $1{\text{m/s}}^2$ until its velocity becomes zero. The total distance covered by the car is -

• A. $1400\text{m}$
• B. $1300\text{m}$
• C. $1200\text{m}$
• D. $1100\text{m}$

Answer 1

The correct option is C $1200\text{m}$

For first $20\text{s}$ when car is accelerated:

$s_1=u_1{t}_1+\frac{1}{2}{a}_1{t}_1^2$

$\Rightarrow s_1=0\times 20+\frac{1}{2}\times 2\times {20}^2=400\text{m}$

Also,

$v_1=u_1+a_1{t}_1=0+2\times 20=40\text{m/s}$

When car is decelerated to rest :

$v_2^2-u_2^2=2a_2{s}_2$

$\Rightarrow 0^2-{40}^2=2(-1)s_2$

$\Rightarrow s_2=800\text{m}$

So, total distance,

$s=s_1+s_2=400+800=1200\text{m}$

Hence, option $\left(C\right)$ is the correct answer.