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lesya692 [45]
3 years ago
15

At t = 0, a particle starts at rest and moves along a line in such a way that, at time t, its acceleration is 24t 2 ft / s2. thr

ough how many feet does the particle move during the first 2 seconds?
Physics
1 answer:
svp [43]3 years ago
5 0
The acceleration of the particle at time t is:
a(t)=24t^2 ft/s^2
The velocity of the particle at time t is given by the integral of the acceleration a(t):
v(t)= \int a(t) \, dt = \int (24 t^2) dt=24 \frac{t^3}{3}=8t^3 ft/s
and the position of the particle at time t is given by the integral of the velocity v(t):
x(t)=\int v(t) = \int (8t^3)=8  \frac{t^4}{4}=2t^4 ft

Assuming the particle starts from position x(0)=0 at t=0, the distance the particle covers in the first t=2 seconds can be found by substituting t=2 s in the equation of x(t):
x(2 s)=2 t^4 = 2 (2s)^4=32 ft
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