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A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.?
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A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist...
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A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist...
Solution:
To solve the problem, we can use the equations of motion. Let's first find the time it takes for the bus to reach the speed of the cyclist.
Step 1: Finding the time for the bus to reach the speed of the cyclist
Initial velocity of the bus (u) = 0
Acceleration of the bus (a) = 2 m/s^2
Final velocity of the bus (v) = 20 m/s
We can use the equation:
v = u + at
where t is the time taken.
Substituting the values, we get:
20 = 0 + 2t
t = 10 s
Therefore, it takes 10 seconds for the bus to reach the speed of the cyclist.
Step 2: Finding the distance covered by the bus in 10 seconds
We can use the equation:
s = ut + 1/2 at^2
where s is the distance covered.
Substituting the values, we get:
s = 0 + 1/2 (2) (10)^2
s = 100 m
Therefore, in 10 seconds, the bus covers a distance of 100 m.
Step 3: Finding the time taken by the cyclist to overtake the bus
Let's assume that the time taken by the cyclist to overtake the bus is t seconds. During this time, the bus also covers a distance of 96 m.
Distance covered by the cyclist = Distance covered by the bus + 96 m
We can use the equation:
s = ut + 1/2 at^2
for both the cyclist and the bus.
For the cyclist:
s = 20t
For the bus:
s = 100 + 0.5 (2) (t - 10)^2
Setting both equations equal to each other, we get:
20t = 100 + 0.5 (2) (t - 10)^2 + 96
Simplifying the equation, we get:
t^2 - 20t - 512 = 0
Solving for t using the quadratic formula, we get:
t = 12 s (rejected) or t = 8 s
Therefore, the cyclist will be able to overtake the bus after 8 seconds.
Answer: A. 8s.
To solve the problem, we can use the equations of motion. Let's first find the time it takes for the bus to reach the speed of the cyclist.
Step 1: Finding the time for the bus to reach the speed of the cyclist
Initial velocity of the bus (u) = 0
Acceleration of the bus (a) = 2 m/s^2
Final velocity of the bus (v) = 20 m/s
We can use the equation:
v = u + at
where t is the time taken.
Substituting the values, we get:
20 = 0 + 2t
t = 10 s
Therefore, it takes 10 seconds for the bus to reach the speed of the cyclist.
Step 2: Finding the distance covered by the bus in 10 seconds
We can use the equation:
s = ut + 1/2 at^2
where s is the distance covered.
Substituting the values, we get:
s = 0 + 1/2 (2) (10)^2
s = 100 m
Therefore, in 10 seconds, the bus covers a distance of 100 m.
Step 3: Finding the time taken by the cyclist to overtake the bus
Let's assume that the time taken by the cyclist to overtake the bus is t seconds. During this time, the bus also covers a distance of 96 m.
Distance covered by the cyclist = Distance covered by the bus + 96 m
We can use the equation:
s = ut + 1/2 at^2
for both the cyclist and the bus.
For the cyclist:
s = 20t
For the bus:
s = 100 + 0.5 (2) (t - 10)^2
Setting both equations equal to each other, we get:
20t = 100 + 0.5 (2) (t - 10)^2 + 96
Simplifying the equation, we get:
t^2 - 20t - 512 = 0
Solving for t using the quadratic formula, we get:
t = 12 s (rejected) or t = 8 s
Therefore, the cyclist will be able to overtake the bus after 8 seconds.
Answer: A. 8s.
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A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.?
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A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.?.
A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A bus starts from rest moving with an acceleration of 2m/s^2.A cyclist,96 m behind the bus starts simultaneously towards the bus at 20m/s.After what time will he be able to overtake the bus:- A.8s. B.10s. C.12s. D.1s.?.
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