Question

A sports 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)?

Answer 1

Step-by-step 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=(v^2-u^2)/(2d)`

`a=(0-(22.22)^2)/(2* 80000)`

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

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

`a=(-0.00308)/(9.8)\ m/s^2`

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

Hence, this is required solution.