Question

The car on the left in the photo and in figure 12-2 moves in a straight line such that for a short time its velocity is defined by v(t) = 3t² + 2t - 4. Determine its position and acceleration when...

a) t = 2 seconds
b) t = 0 seconds
c) t = 1 second
d) t = -1 second

Answer 1

To determine the position and acceleration of the car at various times, integrate the velocity function to find the position and differentiate the velocity function to find the acceleration.

Step-by-step explanation:

The position of the car can be determined by integrating the velocity function, v(t). To find the position at a given time, we integrate v(t) from 0 to the given time. The resulting function, x(t), will give us the position of the car.

To determine the acceleration at a given time, we take the derivative of the velocity function, v(t). Differentiating v(t) will give us a(t), the acceleration of the car at that time.

a) When t = 2 seconds, the position of the car can be found by integrating v(t) from 0 to 2. The acceleration can be determined by taking the derivative of v(t) with respect to t.

b) When t = 0 seconds, the position of the car is 0, since it starts at rest. The acceleration can be found by taking the derivative of v(t) with respect to t.

c) When t = 1 second, the position of the car can be found by integrating v(t) from 0 to 1. The acceleration can be determined by taking the derivative of v(t) with respect to t.

d) When t = -1 second, the position of the car can be found by integrating v(t) from 0 to -1. The acceleration can be determined by taking the derivative of v(t) with respect to t.