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secondary 3 | A Maths
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need help with part b, pls explain too :)
Date Posted: 3 years ago
Views: 185
Part b is actually simply
Speed of car
= Distance / Time
= Length AB / (3/3600) h
Since 1 second is 1/3600 of an hour.
The three seconds it takes the laser gun to detect the position change from A to B is also the same three seconds taken for the car to move from A to B.
Speed of car
= Distance / Time
= Length AB / (3/3600) h
Since 1 second is 1/3600 of an hour.
The three seconds it takes the laser gun to detect the position change from A to B is also the same three seconds taken for the car to move from A to B.
Part a simply uses the cosine rule, going
AB
= sqrt [120^2 + 160^2 - 2 (120) (160) cos 26]
= 74.0696201
~ 74.1 m
Average speed of car, in km/h
= Distance AB (in km) / Time (in hours)
= (74.0696201 / 1000) km / (3/3600) h
= 88.88354412 km/h
~ 88.9 km/h
AB
= sqrt [120^2 + 160^2 - 2 (120) (160) cos 26]
= 74.0696201
~ 74.1 m
Average speed of car, in km/h
= Distance AB (in km) / Time (in hours)
= (74.0696201 / 1000) km / (3/3600) h
= 88.88354412 km/h
~ 88.9 km/h
thx :)
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pls verify...
Date Posted:
3 years ago
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