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 15.0m/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?
Answers
Answered by
2
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.
Similar questions