Question:

A car is traveling along a straight road. The velocity of the car is given by the equation:

v(t) = 3t + 2

Where v(t) is the velocity of the car in meters per second (m/s) and t is the time in seconds (s). Determine:

a) The initial velocity of the car.

b) The average velocity of the car over the time interval from t = 1s to t = 4s.

c) The acceleration of the car.

Answer:

a) The initial velocity of the car can be determined by evaluating the velocity equation at t = 0.

v(0) = 3(0) + 2 = 2 m/s

Therefore, the initial velocity of the car is 2 m/s.

b) The average velocity over a time interval is given by the change in displacement divided by the change in time. In this case, we need to calculate the displacement of the car between t = 1s and t = 4s.

The displacement can be obtained by integrating the velocity equation over the given time interval:

d(t) = ∫ v(t) dt = ∫ (3t + 2) dt

d(t) = (3/2)t^2 + 2t + C

Evaluating the integral limits:

d(4) - d(1) = (3/2)(4)^2 + 2(4) - [(3/2)(1)^2 + 2(1)]

d(4) - d(1) = 24 - 4 = 20 m

The change in displacement is 20 m, and the change in time is 4s - 1s = 3s.

Therefore, the average velocity is:

v_avg = (20 m) / (3 s) = 6.67 m/s

So, the average velocity of the car over the time interval from t = 1s to t = 4s is 6.67 m/s.

c) The acceleration can be obtained by taking the derivative of the velocity equation with respect to time:

a(t) = d(v(t))/dt = d(3t + 2)/dt = 3

Therefore, the acceleration of the car is 3 m/s².