# 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?

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.

Aug 25, 2015

$\text{20.5 m}$

#### Explanation:

I believe it is reasonable to assume average acceleration.

First you need to determine the average acceleration .
Known:
initial velocity, ${v}_{i} = 41 \text{m/s}$
final velocity, ${v}_{f} = 0 \text{m/s}$
time interval, $t = 4.8 \text{s}$

Unknown:
acceleration, $a$

Equation:

$a = \frac{{v}_{f} - {v}_{i}}{t}$

Solution:

$a = \frac{{v}_{f} - {v}_{i}}{t} = \left(0 \text{m/s"-41"m/s")/(4.8"s}\right)$

$a = \text{-8.542 m/s"^2}$

Second you need to determine the velocity at $2.4 \text{s}$. You will need the average acceleration from the first part.
Known:
acceleration, $a = \text{-8.542 m/s"^2}$
beginning velocity, ${v}_{1} = 41 \text{m/s}$
time interval, $t = 2.4 \text{s}$

Unknown:
velocity at $t = 2.4 \text{s}$, which we will represent as ${v}_{2}$

Equation:

$a = \frac{{v}_{2} - {v}_{1}}{t}$

Solution: Rearrange the equation to isolate ${v}_{2}$, then solve for ${v}_{2}$

${v}_{2} = {v}_{1} + a t = 41 \text{m/s"+(-8.542"m/s"^2")(2.4s")="20.5 m}$