Find the distance, in feet, a particle travels in its first 4 seconds of travel, if it moves according to the velocity equation v(t)= −t2 + 4 (in feet/sec).
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To find the distance traveled in the first 4 seconds, we need to integrate the velocity equation with respect to time from 0 to 4.
The velocity equation is given as v(t) = -t^2 + 4.
To integrate the velocity equation, we need to find the antiderivative of v(t).
The antiderivative of -t^2 is -(1/3)t^3, and the antiderivative of 4 is 4t.
Therefore, the antiderivative of v(t) = -t^2 + 4 is -(1/3)t^3 + 4t.
To find the distance traveled, we need to evaluate the antiderivative at the upper limit (4) and subtract the evaluation at the lower limit (0).
So, the distance traveled in the first 4 seconds is:
Distance = [-(1/3)(4)^3 + 4(4)] - [-(1/3)(0)^3 + 4(0)]
= [-(1/3)(64) + 16] - [0 + 0]
= [-64/3 + 16] - [0]
= -64/3 + 48/3
= -16/3
Therefore, the distance traveled in the first 4 seconds is -16/3 feet.
To find the distance traveled by the particle in the first 4 seconds, we need to calculate the area under the velocity curve from time t = 0 to t = 4 seconds.
The velocity equation, v(t) = -t^2 + 4, gives us the rate of change of displacement with respect to time. To find the displacement over a certain time interval, we integrate the velocity function.
To integrate the velocity function v(t) = -t^2 + 4 with respect to time, we can use the power rule of integration. The power rule states that integrating a function of the form x^n gives us (1/(n+1))x^(n+1).
So, integrating the velocity function v(t) = -t^2 + 4 with respect to time, we get:
∫(-t^2 + 4) dt = -((1/3)t^3) + 4t + C,
where C is the constant of integration.
Now, we can find the displacement of the particle over the time interval [0, 4] seconds by evaluating the definite integral:
∫[0,4] (-t^2 + 4) dt = [-((1/3)t^3) + 4t] evaluated from t = 0 to t = 4.
Plugging in the values, we have:
[-((1/3)(4^3)) + 4(4)] - [(-((1/3)(0^3)) + 4(0)].
Calculating this expression, we get:
[-(64/3) + 16] - [(-0 + 0)] = (-64/3 + 16) = 64/3 - 48/3 = 16/3.
Therefore, the particle travels a distance of 16/3 feet in its first 4 seconds of travel.
s(t) = ∫ v(t) dt
= ∫ 4-t^2 dt
= 4t - 1/3 t^3 + C
Since s(0) = 0, C=0, and
s(t) = 4t - 1/3 t^3