Question

Figure shows the velocity time graph of an object a)calculate the acceleration of the object b) Find the distance covered during its motion.

Answer 1

Answer:

Distance covered during the 1st 10 s =areaofΔABP=

2

1

×10×30=150m

Distance covered during the next 20 sec = area of rectangle BPQC =20×30=600m ∴

Distance covered during the last 15 sec =areaofΔCQD=

2

1

×15×30=225m

∴ Total distance covered = 150m+600m+225m=975m

Answer 2

To find :

(a) The acceleration of the object.

(b) the distance covered during its motion.

solution : case 1 : 0 ≤ t ≤ 10s

acceleration = (change in velocity)/(time duration)

= (60 - 10)/(10 - 0)

= 5 m/s²

case 2 : 10s ≤ t ≤ 15s

acceleration = (60 - 60)/(15 - 10) = 0 m/s²

case 3 : 15s ≤ t ≤ 40s

acceleration = (-40 - 60)/(40 - 15) = -4 m/s²

case 4 : 40s to 55s

acceleration = (0 - (-40))/(55 - 40) = 8/3 m/s²

total distance covered by the object = area enclosed by graph

= area of trapezium formed between 0 to 30s + area of triangle formed between 30 to 55 s

= 1/2 (sum of parallel sides) × height + 1/2 × height × base

= 1/2(5 + 30) × 60 + 1/2 × 40 × (55 - 30)

= 35 × 30 + 20 × 15

= 1050 + 300

= 1350 m

Therefore the total distance covered by the object is 1350 m.