An object's velocity after t seconds is v(t)= 38-2t feet per second.
a. How many seconds does it take for the object to come to a stop (velocity = 0)?
b. How far does the car travel during that time?
c. How many seconds does it take the car to travel half the distance in part (b)?
To find the answers to these questions, we need to find the values of t that satisfy the given conditions.
a. To find the time when the object comes to a stop (velocity = 0), we set v(t) = 0 and solve for t.
0 = 38 - 2t
Rearranging the equation:
2t = 38
t = 38/2
t = 19 seconds
b. To find the distance traveled by the object during the time when it comes to a stop, we need to integrate the velocity function over the interval from 0 to t.
The integral of v(t) dt gives us the displacement, which represents the distance traveled.
∫(38 - 2t) dt = [38t - t^2/2] evaluated from 0 to 19
[38 * 19 - (19^2/2)] - [38 * 0 - (0^2/2)]
[722 - (361/2)] - [0 - 0]
722 - (361/2)
722 - 180.5
541.5 feet
c. To find the time it takes for the object to travel half the distance calculated in part b, we divide the distance traveled by 2 and solve for t.
Distance traveled in part b = 541.5 feet
(541.5) / 2 = 270.75 feet
Now we need to set up the equation:
270.75 = 38t - t^2/2
Rearrange the equation:
t^2 - 38t + 541.5 = 0
Now we can solve this quadratic equation using the quadratic formula or factoring techniques.
Using the quadratic formula:
t = (-(-38) ± sqrt((-38)^2 - 4 * 1 * 541.5)) / (2 * 1)
Simplifying:
t = (38 ± sqrt(1444 - 2166)) / 2
t = (38 ± sqrt(-722)) / 2
Since the square root of a negative number is not a real number, there are no real solutions to this equation. Therefore, the object does not travel half the distance in part b.