Question

The initial position of an object is x=95 m. The initial velocity is v=24 m/s.

Given that acceleration is given by a(t) = -4t +1. What is the maximum value of x and v? At what times do they occur?

Answer 1

given that

a(t) = -4t + 1 = dv/dt

v(t) = integration(a(t) dt)

v(t) = -4t²/2 + t + c

v(t) = -2t² + t + c

v(0) = 24 m/s (given)

so

24 = -2*0 + 0 + c

c = 24

v(t) = -2t² + t + 24

for maximum value of v

dv/dt = 0

-4t + 1 = 0

t = 1/4

maximum value of v = (-2*(1/4)² + 1/4 + 24) = 24.125 m/s

v(t) = dx/dt

x(t) = integration (v(t) dt)

x(t) = -2t³/3 + t²/2 + 24*t + c

x(0) = 95 m (given)

x(t) = -2t³/3 + t²/2 + 24*t + 95

for maximum value of x

dx/dt = 0

-2t² + t + 24 = 0

2t² - t - 24 = 0

t = 3.72 sec

maximum value of x = -2*(3.72)³/3 + (3.72)²/2 + 24*(3.72) + 95 = 156.88 m