Question

A car starting from rest accelerates at a constant 2.0 m/s2 for 10 s. It then travels with constant speed it has achieved for another 10 s. Then it finally slows to a stop with constant acceleration of magnitude 2.0 m/s2. How far does it travel after starting?

Answer 1

Answer:

The total distance will be 400 m.

Explanation:

For portion AB:

Acceleration = 2$m/s^2$

t= 10 s

Car start from rest , u=0 m/s

We know that

$S=ut+\dfrac{1}{2}at^2$

$S=0+\dfrac{1}{2}\times 2\times 10^2$

S= 100 m.

For portion BC:

V= u + at

V=0 + 2 x 10

V= 20 m/s

In this portion car moves with constant velocity  20 m/s for 10 s.

So distance S= V x t

S=20 x 10 =200 m.

For portion CD:

The velocity at point C will be 20 m/s

In this portion the final speed of car will be zero because given that at final car come to rest.

So the acceleration will be in the negative direction to stop the car.

We know that

$v^2=u^2-2aS$

$0=20^2-2\times 2\times S$

S=100 m

The total distance AD=AB + BC+ CD

AD=100 +200 + 100 m

AD=400 m.

The total distance will be 400 m.