Question

Akash starts from rest and moves along the x axis with constant acceleration of 5 metre per second square for 8 seconds if then continue with constant velocity.What distance will the car cover in 12 seconds since it started from rest?​

Answer 1

Answer: 480 m

Explanation: using equation of motion

v = u + at

v = 0 + 5*8

v = 40

d = v * t

d = 40 * 12

d = 480 m

Answer 2

Answer:

320 metres

Explanation:

Given:

Initial velocity = u = 0 m/s

Acceleration = a = 5 m/s²

Time = t = 8 seconds

To find:

Distance covered in 12 seconds since it started from rest

It had a uniform acceleration upto 8 seconds and the rest 4 seconds it moved with a constant velocity

We can find the distance covered in 8 seconds using second equation of motion which says:

$s=ut+\frac{1}{2}at^{2}$

Substituting the values,we get:

We shall use time as 8 as it is accelerated only till 8 seconds

$s=0 \times 8 +\frac{1}{2} \times 5 \times 8 \times 8$

$s= 160 \ metres$

The distance covered in 8 seconds with acceleration is 160 metres

Now we have to find final velocity at the end of acceleration which is to be used for constant velocity:

V=u+at

V=0+5×8

V=40 m/s

Time = 4 seconds

Distance = speed×time

Distance = 40×4

Distance = 160 metres

Total distance = 160 metres+160 metres

Total distance = 320 metres

The car will cover a distance of 320 metres in 12 seconds