A person takes 6s to go once around a triangular path of side 12m . What will be the magnitude of the distance covered and his displacement at the end of 9s
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Distance covered in 6 s is :
12 × 3 = 36 m
Since the triangle is equilateral.
Total time = 9s
Total rounds in 9 s = 9/6 = 1.5
Let triangle be ABC
0.5 Indicates that it's half a round hence can get the displacement.
Half of the triangle forms a right angled triangle with hyp = 12, base = 6
We are looking for h which is the displacement.
We will use Pythagoras theorem :
12² - 6² = 108
√108 = 10.39
This is the displacement magnitude.
The distance equals to :
1.5 × 36 = 54m
12 × 3 = 36 m
Since the triangle is equilateral.
Total time = 9s
Total rounds in 9 s = 9/6 = 1.5
Let triangle be ABC
0.5 Indicates that it's half a round hence can get the displacement.
Half of the triangle forms a right angled triangle with hyp = 12, base = 6
We are looking for h which is the displacement.
We will use Pythagoras theorem :
12² - 6² = 108
√108 = 10.39
This is the displacement magnitude.
The distance equals to :
1.5 × 36 = 54m
Answered by
0
Answer:
Explanation:
Distance covered in 6 s is :
12 × 3 = 36 m
Since the triangle is equilateral.
Total time = 9s
Total rounds in 9 s = 9/6 = 1.5
Let triangle be ABC
0.5 Indicates that it's half a round hence can get the displacement.
Half of the triangle forms a right angled triangle with hyp = 12, base = 6
We are looking for h which is the displacement.
We will use Pythagoras theorem :
12² - 6² = 108
√108 = 10.39
This is the displacement magnitude.
The distance equals to :
1.5 × 36 = 54m
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