A driver of car travelling at 52 km h-1 applies the brakes and accelerates uniformly in the opposite direction the car stops in 5 s another driver going at 3 km h-1 in another car applies his brakes slowly versus time graph for the two cars. Which of the two cars travelled farther after the brakes were applied?
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Answered by
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Distance Travelled by first car before coming to rest =Area of △ OPR
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
= (1/2) x OR x OP
= (1/2) x 5 s x 52 kmh-1
= (1/2) x 5 x (52 x 1000) / 3600) m
= (1/2) x 5x (130 / 9) m
= 325 / 9 m
= 36.11 m
Distance Travelled by second car before coming to rest =Area of △ OSQ
= (1/2) x OQ x OS
= (1/2) x 10 s x 3 kmh-1
= (1/2) x 10 x (3 x 1000) / 3600) m
= (1/2) x 10 x (5/6) m
= 5 x (5/6) m
= 25/6 m
= 4.16 m
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Answer:
Here is your answer
Explanation:
(A) : Initial speed of the car A u=52 kmh
−1
=52×
18
5
=14.44 ms
−1
The car stops in 5 seconds i.e v=0 at t=5
(A) : Initial speed of the car A u=3 kmh
−1
=3×
18
5
=0.83 ms
−1
The car stops in 10 seconds i.e v=0 at t=10
With the help of these initial and final points for both the cases, we can plot the graph of speed vs time.
Area under the speed-time graph gives the distance covered.
∴ Distance covered by car A x
A
=
2
1
×14.44×5=36.1 m
∴ Distance covered by car A x
B
=
2
1
×0.83×10=4.15 m
Thus car A travels more distance than B.
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