Question

A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min.

a. What was her average speed?
b. If the straight-line distance from her home to the university is 10.3 km in a direction 25.00 south of east, what was her average velocity?
c. If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?

Answer 1

Answer:

a) $\rm speed =11.11\ m.s^{-1}$

b) $v=9.537\ m.s^{-1}$

c) $\rm speed_{avg}=0.8547\ m.s^{-1}$

   $v=0\ m.s^{-1}$

Explanation:

Given:

• distance from home to university as measured by the odometer, $d=12\ km$

• times taken to cover this distance, $t=18\ min=1080\ s$

a)

average speed:

$\rm speed=\frac{d}{t}$

$\rm speed=\frac{12000}{1080}$

$\rm speed =11.11\ m.s^{-1}$

b)

given that displacement, $s=10.3\ km$

Therefore velocity:

$v=\frac{s}{t}$

$v=\frac{10300}{1080}$

$v=9.537\ m.s^{-1}$

c)

time taken in returning, $t_r=7hr.\ 30min.=27000\ s$

Therefore total time in the round trip, $t_t=27000+1080=28080\ s$

After returning home the total displacement is zero so, average velocity:

$\rm v=\frac{total\ displacement}{total\ time}$

$v=0\ m.s^{-1}$

And the average speed:

$\rm speed_{avg}=\frac{total\ distance}{total\ time}$

$\rm speed_{avg}=\frac{2\times 12000}{28080}$

$\rm speed_{avg}=0.8547\ m.s^{-1}$