Question

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=bt2−ct3, where b = 2.60m/s2 and c = 0.130 m/s3 Calculate the average velocity of the car for the time interval t= 0 to t= 10.0 s.

Answer 1

The average velocity of the car from t = 0 to t = 10.0 s, using the equation x(t)=bt^2-ct^3 with b = 2.60 m/s^2 and c = 0.130 m/s^3, is calculated to be 13 m/s.

Step-by-step explanation:

To calculate the average velocity of the car for the time interval from t = 0 to t = 10.0 s, we can use the given position function x(t) = bt2 - ct3 where b = 2.60 m/s2 and c = 0.130 m/s3. The average velocity is defined as the total displacement divided by the total time taken.

Using the formula for displacement, we first find the position of the car at t = 0 and t = 10.0 s:

• x(0) = b(0)2 - c(0)3 = 0
• x(10.0) = b(10.0)2 - c(10.0)3

Substituting the values of b and c we get:

• x(10.0) = 2.60(10.0)2 - 0.130(10.0)3

Calculating those we find:

• x(10.0) = 2.60(100) - 0.130(1000)
• x(10.0) = 260 - 130 = 130 meters

Now we can find the average velocity vavg:

vavg = displacement / time = x(10.0) - x(0) / 10.0 - 0

vavg = 130 m / 10.0 s = 13 m/s

Therefore, the average velocity of the car over the 10-second interval is 13 m/s.