(a) 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 the first
2 seconds?
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sabrina1234:
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Answered by
4
ANSWERS
>>>>>>>>>>>>>>>
(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
>>>>>>>>>>>>>>>>>>>
HOPE IT HELPS :):):):):):):):):):):):):):)
>>>>>>>>>>>>>>>
(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
>>>>>>>>>>>>>>>>>>>
HOPE IT HELPS :):):):):):):):):):):):):):)
Answered by
3
1. 15m
you posted this question 2nd time
you posted this question 2nd time
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