Question

Q5. A car travels from stop A to stop B with a speed of 30 km/h and then returns back to A with a speed of 50 km/h.

Find

(i) displacement of the car.

(ii) distance travelled by the car.

(iii) average speed of the car.​

Answer 1

Answer:

$\boxed{\textbf{\large{displacement=0 km}}}$

$\boxed{\textbf{\large{Average speed = 37.5 km/hr}}}$

$\boxed{\textbf{\large{total distance = 2 km}}}$

Explanation

A car travels from A to B with a speed 30km/hr

Return back to point A with a speed of 50km/hr

☑ here displacement is Zero,

cause displacement is shortest distance between initial and final point

$\boxed{\textbf{\large{displacement=0 km}}}$

displacement=0 km

☑ To calculate the distance travelled by a car, we have to know speed and time, but there is no time mention in the question,

so, take variable m is a as distance

distance = m

car travels a distance from A to B and then return back to A therefore take a distance 2m

total distance = 2m

let's find time,

we know,

speed = distance /time

☑ the time taken to the car from A to B =time = distance /speed

time = m / 30

☑ from B to A speed of the car is 60km/hr

time = m /50

so, total time =[ m / 30 + m / 50 ]

= [8m /150 ] hr

therefore Now,

☑ Average speed

Average speed = total distance / total time,

= 2m / [8m /150]

=2m x ( 150 /8m)

=37.5 km /hr

Average speed = 37.5 km/hr

$\boxed{\textbf{\large{Average speed = 37.5 km/hr}}}$

☑ total distance

total distance =total time X speed

2m= [8m/150] X 37.5

m= 2/2=1

therefor,

total distance = 2m = 2(1)=2

= 2 km

$\boxed{\textbf{\large{total distance = 2 km}}}$

Answer 2

1) Displacement = s 2- s 1= (50-30)km = 20 km.(As it's in a particular direction)

2) Distance = (30+50)km= 80km(As it's irrespective of the direction)

3) Average Speed of the car = Total distance / Total time taken

=>Average speed = (30+50)/1+1 km/hr.

=>Average speed = 80/2 km/hr.

=>Average speed = 40 Km/hr.

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