Question

a bike initially at rest accelerates at the rate of 2.4m/s^2 for 15s.it then continues to move with the same speed for 20s. it is further brought to rest in another 6s.calculate the maximum velocity reached, retardation and the distance travelled

Answer 1

Answer:

u (initial velocity) = 0m/s

a(acceleration) = 2.4m/s²

t1 = 15s

1st phase, object accelerating at 2.4m/s²

v-u/t = a

v-0/15 = 2.4

v = 36 m/s

Distance travelled = s = 1/2 at²

= 1/2 * 2.4 * 15²

d1= 270m

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2nd phase, object moving at a constant speed of 36m/s

d = s*t

= 36 * 20

d2= 720 m

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3rd phase object coming to rest.

v = 0m/s

u = 36m/s

t = 6s

v = u + at

0 = 36 + 6a

a = -6m/s²

s = ut + 1/2 at²

= 36*6 + 1/2 * -6*36

d3= 108m

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maximum velocity reached = 36m/s

retardation or negative acceleration = -6.0m/s²

distance travelled = d1+d2+d3 = 1098 m

Answer 2

Answer:

maximum velocity reached = 36 m/s

retardation = -6. 0 m/s^2

distance travelled in 3 days = 1098m.

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