Physics, 07.07.2019 18:30 chrisraptorofficial

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?

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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.

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.

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.

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

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