Question

a car moving with constant acceleration covers the distance between two points 58.0m apart in 6.20s it's speed as it passes the second point is 15.0​m/s a. what is the speed at the first point? b. what is the acceleration? c. at what prior distance from the first point was the car at rest?​

Answer 1

Answer:

u≈3.71m/s , a≈1.82m/s² , s≈-156m

Explanation:

since we know that

v=u+at

therefore

15=u+a(6.2)...(1)

s=ut+1/2at²

58=6.2u+1/2a(6.2)²...(2)

from 1. (15-u)/6.2=a

therefore

58=6.2u+1/2[(15-u)/6.2]×6.2²

58=6.2u+1/2(15-u)6.2

58=6.2u+(15-u)3.1

58=6.2u+46.5-3.1u

(58-46.5)/3.1=u

u≈3.71

therefore a=(15-3.71)/6.2≈1.82

since s=ut+1/2at²

when car was on rest

s=1/2(1.82)(6.2)²

s≈156m

hence 156m ago the car was at rest.