Question

A 60kg athlete sprints 40m in 5.0s at a constanr acceleration. What is

surT Determine (a) the speed of the object when it reaches point O. (b) the distance, x. 13. A 60 kg athlete sprints 40 m in 5.0 s at a constant acceleration. What is (b) (c) the magnitude of the horizontal force acting on the athlete? the average power output during the first 3.0 s of her run? the average power output during the final 2.0 s of her run?

Answer 1

Given that he sprints, which means initial velocity, U = 0 m/sec

d = 40 m

t = 5.0 sec

Using 2nd kinematic equation

d = U*t + 0.5*a*t^2

40 = 0*5 + 0.5*a*5^2

a = 40/(0.5*5^2)

a = 3.2 m/sec^2 = acceleration of athelete

Now from Newton's 2nd law:

Fnet = m*a

Fnet = 60*3.2 = 192 N

Part B.

Average power is given by:

P_avg = F*ds/dt

ds = distance traveled in first 3 sec

ds = 0*3 + 0.5*3.2*3^2

ds = 14.4 m

dt = 3 sec

F = 192 N

So,

P_avg = 192*14.4/3 = 921.6 N

Part 3.

ds = distance traveled in final two second = 40 - 14.4 = 25.6 m

dt = 2 sec

So,

P_avg = 192*25.6/2 = 2457.6 N

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