Physics Help?

Question

A car slows down from 112 km/h to 40.0 km/h in 5.00 s. 

a.​What is its average acceleration? Use SI units. 

b.​How much distance does it cover during these 5.00 s?

Answer 1

(a)

At time (t) the speed of the car is 112km/h

After 5 sec ( t+5 ) the speed of the car is 40km/h

The change in velocity is just the difference in the two velocities. So the difference between 112 km/h and 40 km/h can be calculated as

=112-40

=72km/hr

This means that the car used an unknown acceleration to increase its velocity by 72km/hr.

Acceleration is calculated by the formula: a_av = Dv/Dt, so we can put the numbers in and get this

  a= (72km/hr) / (5 sec)

To convert km/h into m/s we have to multiply with 5/18

a = 4 m/s²

Average acceleration of car is a = 4 m/s²

Answer 2

(b)

To find how much distance it travels  in this time, we can use the formula v²=u² + 2as,

where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance. 

Now put the values in :

(112 km/hr)² = (40 km/hr)² + 2 × 4 m/s² ×s

2 × 4 m/s² ×s = 10944 km²/h²

s = 105.556 m

The distance the car can cover in 5 sec is s = 105.556 m