Question

suppose that an accelerating car goes from 0 mph to 54.5 mph in five seconds. its velocity is given in the following table, converted from miles per hour to feet per second, so that all time measurements are in seconds. (note: 1 mph is 22/15 ft/sec.) find the average acceleration of the car over each of the first two seconds.

t (s)	0	1	2	3	4	5
v(t) (ft/s)	0.00	27.27	47.27	61.82	72.73	80.00

average acceleration over the first second =
help (units)
average acceleration over the second second =
help (units)

Explanation:

Step1: Recall average - acceleration formula

The formula for average acceleration $a_{avg}=\frac{\Delta v}{\Delta t}$, where $\Delta v = v_f - v_i$ and $\Delta t$ is the time - interval.

Step2: Calculate average acceleration for the first second

For the first second ($t = 0$ to $t = 1$), $v_i=v(0) = 0.00$ ft/s and $v_f=v(1)=27.27$ ft/s, $\Delta t = 1 - 0=1$ s.
$a_{avg1}=\frac{v(1)-v(0)}{1 - 0}=\frac{27.27 - 0.00}{1}=27.27$ ft/s².

Step3: Calculate average acceleration for the second second

For the second second ($t = 1$ to $t = 2$), $v_i = v(1)=27.27$ ft/s and $v_f=v(2)=47.27$ ft/s, $\Delta t=2 - 1 = 1$ s.
$a_{avg2}=\frac{v(2)-v(1)}{2 - 1}=\frac{47.27 - 27.27}{1}=20$ ft/s².

Answer:

average acceleration over the first second = 27.27 ft/s²
average acceleration over the second second = 20 ft/s²