A sports car go from rest to 32 m/s in 3.88s. The same car can come to a full stop from that speed in 3.96 s. What is the ratio of starting to stopping accelerations? Show your work please.

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llltkl | College Teacher | (Level 3) Valedictorian

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Acceleration is the rate of change of velocity.

`a=(Deltav)/(Deltat)`

For the starting phase:

`a=(32-0)/3.88=32/3.88 =8.247 m/s^2`

For the stopping phase,

`a=(0-32)/3.96=-32/3.96 =-8.0808 m/s^2`

The negative sign indicates that here retardation operates actually.

So, the ratio of starting to stopping acceleration of the sports car is `8.247/-8.0808=-1.02`

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