Question

John drives a distance of 90 Km to the right at a steady speed of V1=25.00 m/s. Then he stops at the gas station for 10 minutes. Then it returns and drives back ( to the left) for half an hour to the left at speed of V2 = 20.0 m/s.Speed =total distance/total timeVelocity=Displacement /total time1 km= 1000 m, 1 minute = 60 seconds, 1 hour= 60 minutesa. What is the difference between speed and Velocity?b. Calculate the total time of his round trip. c. Calculate the total distance.d. Calculate the total displacement.e. Calculate the average speed. f. Calculate the average velocity.

Answer 1

a. Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

b. $t=6000\ s$ is the total time taken in the trip

c. $d=126000\ m$ is the total distance

d. $s=54000\ m$ towards right from the starting point.

e. $v_a=21\ m.s^{-1}$

f. $\vec v_a=9\ m.s^{-1}$ towards right.

Explanation:

a.

Speed is a scalar quantity while velocity is a vector quantity.

Speed is defined as rate of change of distance per unit time whereas velocity is defined as rate of change of displacement per unit time.

Speed is a directionless quantity while velocity constitutes direction.

b.

Total time of round trip when we're given:

• distance travelled to the right, $d_r=90000\ m$

• speed while travelling to the right, $v_r=25\ m.s^{-1}$

• time spent at gas station, $t_g=600\ s$

• time spent while travelling back towards the left, $t_l=30\times 60=1800\ s$

• speed while travelling to the left, $v_{_l}=20\ m.s^{-1}$

Now time taken for travelling towards right:

$t_r=\frac{d_r}{v_r}$

$t_r=\frac{90000}{25}$

$t_r=3600\ s$

Therefore total time taken in the round trip:

$t=t_r+t_l+t_g$

$t=3600+600+1800$

$t=6000\ s$

c.

Now, distance travelled towards left:

$d_l=v_{_l}\times t_l$

$d_l=20\times1800$

$d_l=36000\ m$

Therefore total distance:

$d=d_l+d_r$

$d=36000+90000$

$d=126000\ m$

d.

Now, total displacement:

$s=d_r-d_l$

$s=90000-36000$

$s=54000\ m$ towards right from the starting point.

e.

Average speed:

$v_a=\frac{d}{t}$

$v_a=\frac{126000}{6000}$

$v_a=21\ m.s^{-1}$

f.

Average velocity:

$\vec v_a=\frac{s}{t}$

$\vec v_a=\frac{54000}{6000}$

$\vec v_a=9\ m.s^{-1}$ towards right.