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
Aug 23, 2015

Answer:

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

Aug 25, 2015

Answer:

#"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"#