A resting car starts accelerating uniformly at $a_{1}$ $a_1$ rate a then moves at a constant velocity =4m/s then decelerates at a constant rate $a_{2}$ $a_2$ . the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?

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A resting car starts accelerating uniformly at $a_{1}$ $a_1$ rate a then moves at a constant velocity =4m/s then decelerates at a constant rate $a_{2}$ $a_2$ . the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?

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Answer 1 · 2018-07-21T14:47:33

The total distance was $40 m$ $40 m$ .

Explanation:

The car accelerates uniformly from rest to $4 \frac{m}{s}$ $4 m/s$ over some time $t_{1}$ $t_1$ . Then it decelerates uniformly for another time, $t_{2}$ $t_2$ . So the greatest speed was $4 \frac{m}{s}$ $4 m/s$ . 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 $2 \frac{m}{s}$ $2 m/s$ for 10 s. It would have reached $2 \frac{m}{s}$ $2 m/s$ in a time of $\frac{t_{1}}{2}$ $t_1/2$ . And during the deceleration, it would have decreased to $2 \frac{m}{s}$ $2 m/s$ in a time of $\frac{t_{2}}{2}$ $t_2/2$ . Therefore

$\frac{t_{1}}{2} + \frac{t_{2}}{2} = 10 s and t_{1} + t_{2} = 20 s$ $t_1/2 + t_2/2 = 10 s and t_1 + t_2 = 20 s$

Because both $a_{1} and a_{2}$ $a_1 and a_2$ are constant, and the peak velocity was $4 \frac{m}{s}$ $4 m/s$ , you can verify that the average velocity was $2 \frac{m}{s}$ $2 m/s$ using the formula

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

Therefore the total distance was

$2 \frac{m}{s} \cdot (t_{1} + t_{2}) = 2 \frac{m}{s} \cdot 20 s = 40 m$ $2 m/s * (t_1 + t_2) = 2 m/s * 20 s = 40 m$

I hope this helps,
Steve

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A resting car starts accelerating uniformly at $a_{1}$ $a_1$ rate a then moves at a constant velocity =4m/s then decelerates at a constant rate $a_{2}$ $a_2$ . the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?

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Explanation:

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Math

Humanities

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A resting car starts accelerating uniformly at a1a_1 rate a then moves at a constant velocity =4m/s then decelerates at a constant rate a2a_2. the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?

1 Answer

Explanation:

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A resting car starts accelerating uniformly at $a_{1}$ $a_1$ rate a then moves at a constant velocity =4m/s then decelerates at a constant rate $a_{2}$ $a_2$ . the velocity of the car is greater than 2m/s for 10sec. What is the distance traversed by the car?