Question

A truck stops from a speed of 72 km/h in 40 s. Calculate the stopping distance.

Answer 1

Answer:

uinitial=72×185=20m/s

adec=2m/s2

(a)s=ut+21at2

v=u+at

v=0

u−adect=0

20=2×1

t=10s

s=20×10−21×2×100

s=100m

(b)t=10s

(c)snth=u+2a(2n−1)

s1st=20−22(2×1−1)

s1st=19m

s3rd=20−22(2×3−1)

s3rd

Explanation:

I hope you got your answer

Answer 2

Answer:

uinitial=72×185=20m/s

adec=2m/s2

(a)s=ut+21at2

v=u+at

v=0

u−adect=0

20=2×1

t=10s

s=20×10−21×2×100

s=100m

(b)t=10s

(c)snth=u+2a(2n−1)

s1st=20−22(2×1−1)

s1st=19m

s3rd=20−22(2×3−1)

s3rd

Explanation:

I hope you got your answer