# A resting car starts accelerating uniformly at a_1 rate a then moves at a constant velocity =4m/s then decelerates at a constant rate a_2. the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?

Jul 21, 2018

The total distance was $40 m$.

#### Explanation:

The car accelerates uniformly from rest to $4 \frac{m}{s}$ over some time ${t}_{1}$. Then it decelerates uniformly for another time, ${t}_{2}$. So the greatest speed was $4 \frac{m}{s}$. We do not really know if the story continues until the car comes to a stop.We are asked to find total distance, but over how much of the story? Until the deceleration brings the car to a stop?

I will assume we need to find distance from start until the deceleration brings the car to a stop.

We are also told that the velocity was greater than $2 \frac{m}{s}$ for 10 s. It would have reached $2 \frac{m}{s}$ in a time of ${t}_{1} / 2$. And during the deceleration, it would have decreased to $2 \frac{m}{s}$ in a time of ${t}_{2} / 2$. Therefore

${t}_{1} / 2 + {t}_{2} / 2 = 10 s \mathmr{and} {t}_{1} + {t}_{2} = 20 s$

Because both ${a}_{1} \mathmr{and} {a}_{2}$ are constant, and the peak velocity was $4 \frac{m}{s}$, you can verify that the average velocity was $2 \frac{m}{s}$ using the formula

#v_"ave" = (u + v)/2

Therefore the total distance was

$2 \frac{m}{s} \cdot \left({t}_{1} + {t}_{2}\right) = 2 \frac{m}{s} \cdot 20 s = 40 m$

I hope this helps,
Steve