Question

An automobile moving along a straight track changes its velocity uniformly from 40.0
m/s to 80.0 m/s in a distance of 200 m. What is the acceleration of the vehicle during
this time?

Answer 1

Answer:

12 m/s ( square)

Explanation:

initial position= 40 m/s

final position= 80 m/s

distance= 200 m

$2as = v {}^{2} - u {}^{2}$

here, a = acceleration

$2 \times a \times 200 = 80 {}^{2} - 40 {}^{2}$

400a = 6400 - 1600

400a = 4800

a = 4800 ÷ 400

a= 12

acceleration =

$12metre \: per \: second {}^{2}$