Physics, 28.09.2019 16:30 yerranne

Asports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. what is its acceleration and how many g's is this (g=9.8 m/s^2)?

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Answer from: simrankaurdhatt

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car. It can be calculated using third equation of motion as :

v^2-u^2=2ad

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

$a=\dfrac{0-(22.22)^2}{2\times 80000}$

$a=-0.00308\ m/s^2$

Value of g, $g=9.8\ m/s^2$

$a=\dfrac{-0.00308}{9.8}\ m/s^2$

$a=(-0.000314)\ g\ m/s^2$

Hence, this is required solution.

Answer from: Quest

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Asports car is advertised to be able to stop in a distance of 50.0 m from a speed of 80 km. what is...

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