Question

A car accelerates form rest to a speed of 42m/s in 8s.
Find
1. acceleration of the car
2. Find the distance the car travels in first 8s
3. What is the speed of the car 10s after it begins its motion​

Answer 1

Answer:

• Acceleration of the car = 5.25m/s²

• distance the car travels in first 8s = 168m

• the speed of the car 10s after it begins its motion = 52.5m/s

For step by step solution you can refer to the attachment.

hope it helps you✌

Answer 2

We have,

Initial velocity = 0 m/s

Final speed = 42 m/s

Time = 8 s

We know,

$\boxed{a = \frac{∆v}{∆t} }$

⇒ a = (42 m/s)/(8 s)

⇒ a = 5.25 m/s² (1)

We know,

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

⇒ S = ½ at²

⇒ S = ½ (5.25 m/s²)(8 s)²

⇒ S = 32 × 5.25 m

⇒ S = 168 m (2)

Also,

$\boxed{v = u + at}$

⇒ v = at = (5.25 m/s²)(10 s)

⇒ v = 52.5 m/s (3).