A particle moves according to a law of motion s = f(t), t > = 0, where t is meas

Question

A particle moves according to a law of motion s = f(t), t > = 0, where t is measured in seconds and s in feet. f(t)=t^3-15t^2+72t (a) Find the velocity at time t. (b) What is the velocity after 5 s? (c ) When is the particle at rest? (d) When is the particle moving in the positive direction? (Enter your answer in interval notation.) (e) Find the total distance traveled during the first 7 s. (f) Find the acceleration at time t. Find the acceleration after 5 s. (g) Graph the position, velocity, and acceleration functions for the first 7 s. (h) When, for 0

Explanation / Answer

S = f(t) = t3 - 15t2 + 72t

a). velocity = ds/dt = 3t2 - 30t + 72

b). velocity after t = 5 sec, velocity = 3(5)2 - 30x5+ 72 = 75 - 150 + 72 = 147 - 150 = -3 ft/sec

c). when particle at rest, velocity = ds/dt = 3t2 - 30t + 72 = 0

t2 - 10t + 24 = 0

t2 - 6t - 4t + 24 = 0

(t - 6)(t - 4) = 0

t = 6 sec or 4 sec

d). positive interval notation, 4<= t <= 6

e). total distance during t = 7 sec

S = f(t) = t3 - 15t2 + 72t = 73 - 15(7)2 + 72x7 = 112 ft

f). aceleration = d^2s/dt^2 = 6t - 30

after 5 sec, aceleration = 6x5 - 30 = 0

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