Question

A car starts from rest and accelerates at 1 $m{s}^{-2}$ for 4 s. It then moves at a constant velocity for next 10 s.then decelerates at 2.0 $m{s}^{-2}$ adn finally comes to rest.The total distance travelled by the car is

• A. 52 m
• B. 40 m
• C. 48 m
• D. 46 m

Answer 1

The correct option is A 52 m

During first $4s$ the distance traveled, $x_1=ut+a{t}^2/2=0+\frac{1\times 16}{2}=8m$

At the end of these $4s$ it will gain velocity $v=at=1\times 4=4m/s$

So the new distance in next $10sec$ with constant velocity $v$

is $x_2=v\times 10=40m$

and then it comes to rest after travelling a distance $s$ with retardation $a_1$ so this $s=\frac{v^2}{2a_1}=\frac{16}{2\times 2}=4m$

Total distance is $d=x_1+x_2+s=8+40+4=52m$

Option A is correct.