What is the position of the car when t = 20?

Question

A car travels along a straight road for 30 seconds starting at time t = 0. Its acceleration in ft/sec² is given by the linear graph below for the time interval [0, 30]. At t = 0, the velocity of the car is 0 and its position is 10.

What is the position of the car when t = 20? You must show your work answer to 3 decimal places. Include units in your answer.

Answer 1

The acceleration is a = -x + 10

Found the slope using points (0,10) and (10,0)

(0-10)/(10-0) = -1

y intercept is 10.

Velocity is integral (-x + 10) dx = -x^2/ 2 + 10x + c

One value for that is given. When x = 0, the velocity is 0.

0 = -0^2 /2 + 10(0) + c = 0 + 0 + c = c

c = 0. v = -x^2 / 2 + 10x

Integral (-x^2 / 2 + 10x) = position = -x^3 / 6 + 5x^2 + c

At x = 0 the position is 10. Plug it into the formula to find c.

-0^3 / 6 + 5(0)^2 + c = 10 so c=10

Position = -x^3 / 6 + 5x^2 + 10

At x=20 (t=20, but we're using x as the variable) P = -20^3 / 6 + 5(20)^2 + 10 =

-8000/6 + 2000 + 10 = -8000/6 + 12000/6 + 60/6 = 4060/6 = 676.67 feet