Question

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

The distance travelled by the car in first 4 seconds:

$x_1=ut+\frac{1}{2}a{t}^2$

$x_1=0\left(t\right)+\frac{1}{2}\times 1\times (4{)}^2$

$x_1=8m......Eq:01$

The velocity (v) at the end of these 4 seconds with constant velocity is:

$v=u+at$

$v=0+\left(1\right)\left(4\right)$

$v=4m{s}^{-1}$

New distance $x_2$ in the next 10 seconds will be:

$x_2=v\times 10$

$x_2=4\times 10=40m......Eq:02$

Use formula to find the distance when the car started retarded:

$v^2-u^2=2a\left(x_3\right)$

$(4{)}^2-0^2=2(2)\times (x_3)$

$16=4\times \left(x_3\right)$

$\left(x_3\right)=4m......Eq:03$

Required total distance travelled:

Adding Eq:01; Eq:02 and Eq:03;

$d=x_1+x_2+x_3$

$d=(8+40+4)m$

$d=52m$

OPTION (A) IS CORRECT.