Question

a car travels from stop A to stop B with a speed of 30km/h and then returns back to A with a speed of 50km/h . find
i) displacement of the car
ii) distance travelled by the car
iii) average speed of the car.​

Answer 1

Explanation:

II)AVG SPEED OF CAR = 30+50=80KMH

FIRST ONE HOUR =30

SECOND ONE HOUR=50

D=S×T=80×2=160KM

III)AVG SPEED OF CAR=30+50=80KMH

Answer 2

AnswEr :

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

We have to find i) Displacement of car ii) Distance travelled by car iii) Average speed of car.

Let the distance from A to B be d km.

As car goes from stop A to stop B and returns back to stop A from stop B, that means the car has traveled the same distance twice.

Now as the car comes back to its initial position i.e. stop A, so displacement = 0

→ Total displacement = AB - BA

→ Total displacement = d - d

→ Total displacement = 0

∴ i) Displacement of car = 0 [Answer]

___________________________

Now as the car covers the same distance twice,

→ Total distance = AB + BA

→ Total distance = d + d

→ Total distance = 2d km.

∴ ii) Distance travelled by car = 2d km [Ans.]

__________________________

Now we know that,

Distance = Speed × Time

→ d = 30 × Time

→ Time1 = d/30 h

Now time in 2nd step :

→ Time = Distance/Speed

→ Time2 = d/50 h

Now we will use the formula :

Average speed = Total distance/Total time

→ Av. speed = (d + d)/(d/30 + d/50)

→ Av. speed = 2d/[ (5d + 3d)/ 150 ]

→ Av. speed = 2d/[8d/150]

→ Av. speed = 2d × 150/8d

→ Av. speed = 150/4

→ Av. speed = 37.5 km/h

∴ iii) Average speed of car = 37.5 km/h [Ans.]