Question

The velocity time graph of a car is given in the following figure the mass of the car is 1000 kg what is the distance travelled by a car in the first to second what is the braking force applied at end of 5 second to the car to stop with the within one second

Answer 1

Answer:

Explanation:ANSWERS

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(i) The distance travelled = Area under the graph

=> Distance in first two seconds = Area of triangle ABE

=> Distance = 1/2 × 2 × 15

=> 15 m

(ii) Acceleration = slope of graph

=> Acceleration = 15 / (6-5)

=> Acceleration = 15/1

= 15 m/s^2

Force = Mass × Acceleration

Force = 1000 kg × 15 m/s ^2

Force = 15000 N

(iii) Distance = Area under graph

Distance = Area of trapezium

=> Area of trapezium = 1/2 × sum of parallel sides × Distance between them

=> Distance (s) = 1/2 ( BC + AD ) × (CF)

=> s = 1/2 × ( v + u ) × t

v = u + at

=> s = 1/2 ×( u + at + u ) × t

=> s = 1/2 × (2u + at) × t

=> s = 1/2 × 2ut × at^2

s = ut + 1/2 at ^2

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Answer 2

Answer:

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