a car brakes to a halt in 5 seconds. The position of the car versus time is given by:
x = 10t - t^2
a) What is the average speed between 0 s. and 4 s.?
To find the average speed between 0 s and 4 s, we need to calculate the total distance traveled by the car during that time interval and then divide it by the elapsed time.
The distance traveled by the car can be determined by calculating the definite integral of the velocity function over the given time interval. In this case, the velocity function is the derivative of the position function.
First, let's find the velocity function by taking the derivative of the position function:
v(t) = d/dt (10t - t^2)
v(t) = 10 - 2t
Now, we can calculate the distance traveled by integrating the absolute value of the velocity function over the interval [0, 4]:
distance = ∫ |v(t)| dt from 0 to 4
Plugging in the expression for v(t), we get:
distance = ∫ |10 - 2t| dt from 0 to 4
To evaluate this integral, we split it into two cases based on the function inside the absolute value:
Case 1: When 10 - 2t is positive (10 - 2t > 0)
The integral becomes:
∫ (10 - 2t) dt from 0 to 4
Integrating this expression gives us:
∫ (10t - 2t^2/2) dt from 0 to 4
= [5t^2 - t^3] from 0 to 4
= [5(4)^2 - (4)^3] - [5(0)^2 - (0)^3]
= 80 - 64
= 16
Case 2: When 10 - 2t is negative (10 - 2t < 0)
The integral becomes:
∫ (2t - 10) dt from 0 to 4
Integrating this expression gives us:
∫ (2t - 10) dt from 0 to 4
= [t^2 - 10t] from 0 to 4
= [4^2 - 10(4)] - [0^2 - 10(0)]
= 16 - 40
= -24
Since we're calculating the distance, we take the absolute value:
|distance| = |-24| = 24
Now, to find the average speed, we divide the distance by the elapsed time:
average speed = |distance| / elapsed time
= 24 / (4 - 0)
= 24 / 4
= 6 m/s
Therefore, the average speed between 0 s and 4 s is 6 m/s.