Question

A motorcycle is stopped at a traffic light. When the light turns green, the motorcycle accelerates to a speed of 71 km/h over a distance of 50 m.

a) What was the average acceleration of the motorcycle over this distance?

(b) Assuming the motorcycle maintained a constant acceleration, how far is it from the traffic light after 3.1 s?

Answer 1

(a)    required equation is

v^2=u^2+2*a*s

u=0 m/s

v=71 km/h=(71*1000)/3600=19.72 m/s

s=50m

a=((v^2)-(u^2))/2*s

a= ((19.72^(2))-(0))/(2*50)

a=3.88 m/s^2

(b) given that a is constant& t=3.1s

so

s=u*t

s=19.72*3.1

s=61.13 m