Question

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:

Displacement is the shortest distance between the initial and the final position. Here the car travels from stop A to B and return to stop A. Thus, the displacement of the car is 0 m/s.

Explanation:

let the distance travelled by the car from stop A to B = x km. The car travel from A to B return back to B. Thus, total distance travelled = 2x km. Time while going from A to B = distance travelled / speed taken = x / 30 hours. Time while returning from B to A = x / 50 hours . Total time x / 30 + x / 50 = 8x/ 150

average speed = total distance / time taken

2x × 75 / 4x

average speed = 37.5 km/h

thus, total distance = 2x = 2 × 37.5 = 75 m

Answer 2

Answer:

Displacement is the shortest distance between the initial position and the final position. Here, the car travels from stop A to stop B and then returns back to stop A.

Thus, the displacement of the car is 0 m/s.

Let the distance travelled by the car from stop A to stop B = x km.

The car travels from stop A to stop B and returns back to point A.

∴ Total distance travelled = 2x km.

Time taken while going from stop A to stop B = Distance travelled / Speed taken

= x/30 hrs

Time taken while returning from stop B to stop A = Distance travelled / Speed taken.

= x/50 hrs.

Total time taken = x/30 + x/50

                         = (50x + 30x)/1500

                         = 80x/1500

                         = 8x/ 150

                         = 4x/75 hrs

Average speed = Total distance traveled/ total time taken.

                        = 2x × 75/4x

                        = 150x/4x

                        = 37.5 km/h.

Hope this helps :)

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