Question

A car driving at 41m/s hits the brakes and comes to a stop in 4.8 seconds. How fast was it going after 2.4 seconds?

Answer 1

In general it is not possible to answer this question without knowing the form of the deceleration curve.

Explanation:

In the (extremely unlikely) case that deceleration is linear, the car would have been going 20.5 m/s after 2.4 sec.

The image below shows 3 possible (of infinitely many) deceleration patterns with different speeds for each one at time 2.4 seconds.

Answer 2

`"20.5 m"`

Explanation:

I believe it is reasonable to assume average acceleration.

First you need to determine the average acceleration .
Known:
initial velocity, `v_i=41 "m/s"`
final velocity, `v_f=0 "m/s"`
time interval, `t=4.8 "s"`

Unknown:
acceleration, `a`

Equation:

`a=(v_f-v_i)/t`

Solution:

`a=(v_f-v_i)/t=(0"m/s"-41"m/s")/(4.8"s")`

`a="-8.542 m/s"^2"`

Second you need to determine the velocity at `2.4 "s"`. You will need the average acceleration from the first part.
Known:
acceleration, `a="-8.542 m/s"^2"`
beginning velocity, `v_1=41 "m/s"`
time interval, `t=2.4 "s"`

Unknown:
velocity at `t=2.4 "s"`, which we will represent as `v_2`

Equation:

`a=(v_2-v_1)/t`

Solution: Rearrange the equation to isolate `v_2`, then solve for `v_2`

`v_2=v_1+at=41"m/s"+(-8.542"m/s"^2")(2.4s")="20.5 m"`