Question

A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. What is the distance covered before the car comes to a stop? (Round your answer to one decimal place.) ft

Answer 1

Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, $a=-44\ ft/s^2$

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}$

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

Answer 2

Answer:

The distance covered by car before stopping is 61.13 ft.

Explanation:

Given data:

Initial Speed of car is, $u=50 \;\rm mi/h = 50 \times 1.467 =73.35 \;\rm ft/s$.

Deceleration of car is, $a=-44\;\rm ft/s^{2}$. (Negative sign shows negative acceleration)

Applying the second kinematic equation of motion as,

$v^{2}=u^{2}+2as$

Here, s is the distance covered and v is the final speed. Since, car stops finally, v = 0.

Solving as,

$0^{2}=73.35^{2}+2(-44) \times s\\88s =73.35^{2}\\s \approx 61.1 \;\rm ft$

Thus, distance covered by car before stopping is 61.1 ft.

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