An electric vehicle starts from rest and accelerates at a rate of 2.0 m/s^2 in a straight line until it reaches a speed of 20 m/s. The vehicle then slows at a constant rate of 1.0 m/s^{2} until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle travel from start to stop?

#### Solution:

### a) For this part, we will use the equation, v=v_0+at (where, v is velocity final, v_0 is velocity initial, a is acceleration, and t is time). We know that the final speed of the car is 20 m/s, and a =2.0 m/s^2. We also know that velocity initial, v_0, is equal to 0 because the car starts from rest. Now, we simply plug these values into our equation and find the time it took for the car to accelerate to it’s maximum speed. (Note we are only finding the time to the point at which the speed of the car is 20 m/s.) So we have:

### 20=0+2.0t_1

### t_1=10 s.

### Now we will calculate the time it took for the car to decelerate from 20 m/s to 0 m/s at a constant deceleration of 1.0 m/s^2. We will use the same equation, but this time, the velocity initial is 20 m/s, velocity final is 0 m/s and the acceleration is -1.0 m/s^2. Plugging these values in, we have:

### 0=20+(-1.0)t_2

### t_2=20 s.

### Therefore, the total time it took for the car to reachÂ 20 m/s and then come to a stop is 10 s + 20 s=30 s.

### b) To figure out the distance the car travels, we will use the following equation:

### v^2=v_0^2+2a(x-x_0)

### Isolating for x gives us :

### x=\frac{v^2-v_0^2}{2a}

### x=\frac{(20 m/s)^{2}-(0)^{2}}{2(2.0 m/s^{2}}

### x=100 m.

### This only gives us the distance traveled by the car until it reaches a speed of 20 m/s. We will now find the distance the car traveled from 20 m/s to a complete stop using the same equation. Thus, we have:

### x=\frac{(0)^{2}-(20 m/s)^{2}}{2(-1.0 m/s^{2}}

### x=200 m

### Now, to find the total distance traveled, we add the two distances up, 100 m+200 m=300 m.

###### This question can be found in Fundamentals of Physics, 10th edition, chapter 2, question 25.