Your car is initially at rest when your hit that gas and the car begins to accelerate. The forward force on the car is 2830 N while the resisting forces total 130 N. The car has a mass of 1830 kg and the acceleration lasts for 2.8 s. What is the final speed of the car and how much ground does it cover during this acceleration?

A)

The speed of the car at t = 4 s. will be the area under the graph from 0 to 4 s., because the anti derivative of acceleration is velocity and at t = 0 s., v = 0 m/s. So, we have a rectangle with sides of 3 x 4, which gives us 12 m/s as the speed at t = 4 s.

B)

To find the distance traveled we can use simple motion equations. Since there is a constant acceleration of 3 m/s^2 from t = 1 to t = 4 s., the equation s = v (initial) * t + 1/2 * a * t^2 can be used and if we plug in t = 3 s. and v (initial) = 0 m/s, the result is: 13.5 m.

C)

Since from 10 s. to 14 s. there is no acceleration all we need to find is the velocity at t = 10 s. and multiply by 4 seconds, because for an acceleration equal to zero s = v*t. To find the velocity at t = 10, we need to find the total area of the graph from 0 to 10 s. Knowing that the initial velocity at t = 0 s. is 0 m/s, we don't have to add anything to the calculated area. The calculated area is: 3*4 - 1*1 + 3*2 - 2*1 - 1*1 = 14 m/s. Plugging back in into the equation: the distance traveled is: 14 * 4 = 56 m.

D)

To find average speed, we divide distance traveled by time traveled. In a similar way as previously we find that the velocity at t = 6 s. is 11 m/s. And we apply the equation twice, s = v (initial) * t + 1/2 * a * t^2. First for the times between 6 and 8 s. and we get 28 m. Then we apply the equation to the part between 8 and 9 s. (Remember that in this case the acceleration is negative) and the velocity at t = 8 s. is 17 m/s. That value will be used as the initial velocity and after plugging in the numbers the distance on the second part is: 16 m. We add the two distances and get 44 m. and divide is over 3 s. to get: 11.3333333 m/s