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?

2 Answers
Alan P.
Aug 23, 2015

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.
enter image source here

Meave60
Aug 25, 2015

"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"

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