#2 Athlete. A particle of mass m = 0.3 kg is acted upon by a force so that its p

Question

#2 Athlete.

A particle of mass m = 0.3 kg is acted upon by a force so that its position vector is r bar(t) = 2.0t^2 i cap - 5.0 t^2 i cap + t^2 k cap, where t is in s. What is the (instantaneous) power delivered by this force to the particle at time t = 2.0 s? 62 W 72 W 81 W 36 W 70 W An athlete runs the first three laps of a 1600 m race (consisting of 4 laps of 400 m) at a constant speed. On completing the third lap, he finds his time is 3 min, 2.0 s. He immediately accelerates at 0.10 m/s^2 for 6.0 s, and then maintains this new speed to the finish. What is his time for the 1600 m? 4 min, 2.7 s 4 min, 3.6 s 3 min, 57.6 s 3 min, 58.1 s 3 min, 57.9 s A block of mass m_1 = 10.0 kg, sits on a plane inclined an angle theta = 30.0 degree to the horizontal. The coefficient of static friction for the inclined plane is mu_s = 0.40, while the coefficient of kinetic friction is mu_k = 0.30. The block is attached by a massless string passing over a

Explanation / Answer

During first 3 laps:

Distance = 400*3 = 1200m
time = 3*60 + 2 = 182 s
So, v1 = 1200/182 = 6.5934 m/s.

Now, acceleration for 6 sec = .1 m/s^2

Distance covered during 6 s while accelerating:

s = ut + 1/2 at^2 = 6.5934 * 6 + 0.5* 0.1 * 36 = 41.3604 m
Remaining distance = 400 - 41.3604 = 358.6396
This distance he will cover with a speed of v2
v2 = v1 + at = 6.5934 + 0.1*6 = 7.1934 m/s

So, time taken to cover 358.6396 m is = 358.6396/7.1934 = 49.8568 s

So, Total time = 182 + 6 + 49.8568 = 237.8568 s = 3 min 57.9 s FINAL ANS

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