Physics, asked by sabrina1234, 1 year ago

(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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Answers

Answered by flower161
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 :):):):):):):):):):):):):):)


sabrina1234: Thanks
Answered by Anonymous
3
1. 15m
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