A car traveling with velocity v is decelerated by a constant acceleration of magnitude a. It takes a time t to come to rest. If its initial velocity were doubled, the time required to stop would

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Answer:

If the initial speed is doubled the time is also doubled

Explanation:

You have that a car with velocity v is decelerated by a constant acceleration a in a time t.

You use the following equation to establish the previous situation:

[tex]v'=v-at[/tex]     (1)

v': final speed of the car  = 0m/s

v: initial speed of the car

From the equation (1) you solve for t and obtain:

[tex]t=\frac{v-v'}{a}=\frac{v}{a}[/tex]     (2)

To find the new time that car takesto stop with the new initial velocity you use again the equation (1), as follow:

[tex]v'=v_1-at'[/tex]     (3)

v' = 0m/s

v1: new initial speed = 2v

t': new time

You solve the equation (3) for t':

[tex]0=2v-at'\\\\t'=\frac{2v}{a}=2t[/tex]

If the initial speed is doubled the time is also doubled