Question

1. A car travels from a point A to point B at constant speed of 50 km/h. On return
journey along the same path, it travels at a constant speed of 40 km/h. Find its
distance travelled, displacement and average speed.

plss help...tomorrow is my exam.....​

Answer 1

Solution :-

From point A to B

• Speed (S 1 ) = 50 km / hr

• Let the time taken by the car  be t 1 and the distance covered be d

we know that ,

➨ S 1  = d  / t 1

On rearranging ,

➨ t 1  =d / S1  

➨ t 1 = d / 50

From point B to A

• Speed (S 2 ) = 40 km / hr

• Let the time taken by the car  be t 2 and the distance covered be d

we know that ,

➨ S 2 = d  / t 1

On rearranging ,

➨ t 2  =d / S2  

➨ t 2 = d / 40

(a)

➨ Total distance = from A to B  + from B to A

➨ Total distance = d + d

➨ Total distance = 2d

(b)

➨ Displacement = 0

∵ The initial and the final position of the car is the same

(c)

$\implies\mathtt{ Avg.speed =\dfrac{ total \:distance}{ total\: time}}$

$\implies\mathtt{ Avg.speed =\dfrac{ 2d}{ t_1 + t_2 }}$

$\implies\mathtt{ Avg.speed =\dfrac{ 2d}{ \dfrac{d}{50} + \dfrac{d}{40} }}$

$\implies\mathtt{ Avg.speed =\dfrac{ 2d}{d(\dfrac{1}{50} + \dfrac{1}{40} ) }}$

$\implies\mathtt{ Avg.speed =\dfrac{ 2}{\dfrac{4+5}{200} }}$

$\implies\mathtt{ Avg.speed =\dfrac{ 400}{9 }}$

$\implies\mathtt{ Avg.speed =44.4 \dfrac{m}{s }}$

The average speed of the car = 44.4m /s