Question

The velocity time graph of a car is given below. The car weighs 1000 kg.
(i) What is the distance travelled by the car in first 2 seconds?
(ii) What is the braking force at the end of 5 seconds to bring the car to a stop within one second?
b) Derive the equation S=ut+1/2 at² using graphical method.

Answer 1

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

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(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

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(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

Explanation:

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