The velocity of a car after t seconds is 75 – 3t^2 feet per second.
a. How many seconds does it take for the car to come to a stop (velocity = 0)?
b. How far does the car travel while coming to a stop?
c. How many seconds does it take the car to travel half the distance in part (b)?
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a. To find when the car comes to a stop, we need to set the velocity equation equal to 0 and solve for t:
0 = 75 - 3t^2
Rearranging the equation:
3t^2 = 75
Dividing both sides by 3:
t^2 = 25.
Taking the square root of both sides gives us two solutions:
t = ±5.
Since we are interested in positive time, the car comes to a stop after 5 seconds.
b. To find how far the car travels while coming to a stop, we need to find the displacement. The displacement is given by the integral of the velocity function:
∫(75 - 3t^2) dt.
Integrating with respect to t:
∫75 dt - ∫3t^2 dt.
Evaluating the integrals:
75t - t^3 + C.
To find the displacement, we evaluate the expression at t = 5 (the time it takes for the car to come to a stop) and t = 0 (initial time):
Displacement = 75(5) - 5^3 - (75(0) - 0^3) = 375 - 125 = 250 feet.
c. To find the time it takes for the car to travel half the distance in part (b), we divide the displacement in part (b) by 2 and solve for t:
125 = 75t - t^3.
Rearranging the equation:
t^3 - 75t + 125 = 0.
Unfortunately, this equation does not have a simple algebraic solution. We can use numerical methods or graphical methods to approximate the value of t.
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