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iris [78.8K]
3 years ago
6

Determine the stopping distances for a car with an initial speed of 88 km/h and human reaction time of 2.0 s for the following a

ccelerations. (a) a = -4.0 m/s^2 (b) a = -8.0 m/s^2
Physics
1 answer:
seropon [69]3 years ago
4 0

Explanation:

Given that,

Initial speed of the car, u = 88 km/h = 24.44 m/s

Reaction time, t = 2 s

Distance covered during this time, d=24.44\times 2=48.88\ m

(a) Acceleration, a=-4\ m/s^2

We need to find the stopping distance, v = 0. It can be calculated using the third equation of motion as :

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

s=\dfrac{-(24.44)^2}{2\times -4}

s = 74.66 meters

s = 74.66 + 48.88 = 123.54 meters

(b) Acceleration, a=-8\ m/s^2

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

s=\dfrac{-(24.44)^2}{2\times -8}

s = 37.33 meters

s = 37.33 + 48.88 = 86.21 meters

Hence, this is the required solution.

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