Physics, 17.07.2019 12:30 jaleelbrown80

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. through how many feet does the particle move during the first 2 seconds?

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Answer from: mikk33

The distance in feet covered by this particle during the first 2 seconds is 32 feet.

Given the following data:

Initial time = 0 secondsAcceleration = $24t^2 \;ft/s^2$

To calculate the distance in feet covered by this particle during the first 2 seconds:

First of all, we would determine the velocity at which the particle is moving by integrating the value of its acceleration as follows;

$V(t) = \int\limits {(a)} \, dt \\\\V(t) = \int {(24t^2)} \, dt\\\\V(t) = 24 \frac{t^3}{3} \\\\V(t) = 8t^3\; ft/s$

Next, we would calculate the position of this particle by integrating the velocity with respect to time;

$x(t) = \int\limits {(V)} \, dt \\\\x(t) = \int {(8t^3)} \, dt\\\\V(t) = 8 \frac{t^4}{4} \\\\x(t) = 2t^4\; ft$

Now, we can calculate the distance covered by this particle during the first 2 seconds:

$Distance = 2\times 2^4\\\\Distance = 2\times 16$

Distance = 32 feet.

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Answer from: SearchLight

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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