On a dry road, a car with good tires may be able to brake with a constant deceleration of 4.92 m/s^2. (a) How long does such a car, initially traveling at 24.6 m/s, take to stop? (b) How far does it travel in this time?

#### Solution:

### To solve this question, we will use the following equation:

### v=v_0+at where v is final velocity, v_0 is initial velocity, a is acceleration and t is time.

### a)We will now substitute the values given to us in the question. Note that v is equal to 0 because the car comes to a complete stop and the acceleration is negative because it’s deceleration.

### 0=24.6m/s+(-4.92m/s^2)t

### t=\frac{24.6m/s}{-4.92m/s^2}

### t=5.00 s

### Therefore, the time it took for the car to stop is 5 seconds.