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A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:
- a)√110 s
- b)√120 s
- c)10 √2 s
- d)15 s
Correct answer is option 'D'. Can you explain this answer?
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A car is standing 200 m behind a bus, which is also at rest. The two s...
Let's denote the initial position of the car as x_car(0) and the initial position of the bus as x_bus(0). We are given that x_car(0) = x_bus(0) + 200 m.
The position of the car at any time t is given by:
x_car(t) = x_car(0) + v_car(0)t + (1/2)a_car*t^2
Similarly, the position of the bus at any time t is given by:
x_bus(t) = x_bus(0) + v_bus(0)t + (1/2)a_bus*t^2
Since the bus is initially at rest, v_bus(0) = 0. We are also given that a_bus = 2 m/s^2 and a_car = 4 m/s^2.
Substituting the initial positions, we have:
x_car(t) = x_bus(0) + 200 + v_car(0)t + (1/2)a_car*t^2
x_bus(t) = x_bus(0) + (1/2)a_bus*t^2
Since the car catches up with the bus, their positions will be equal at the time of catching up. Therefore, we can set x_car(t) = x_bus(t):
x_bus(0) + (1/2)a_bus*t^2 = x_bus(0) + 200 + v_car(0)t + (1/2)a_car*t^2
Simplifying the equation, we have:
(1/2)a_bus*t^2 - (1/2)a_car*t^2 - v_car(0)t - 200 = 0
Since we are looking for the time at which the car catches up with the bus, we can solve this quadratic equation to find t. However, we need more information about the initial velocity of the car, v_car(0), in order to solve the equation.
The position of the car at any time t is given by:
x_car(t) = x_car(0) + v_car(0)t + (1/2)a_car*t^2
Similarly, the position of the bus at any time t is given by:
x_bus(t) = x_bus(0) + v_bus(0)t + (1/2)a_bus*t^2
Since the bus is initially at rest, v_bus(0) = 0. We are also given that a_bus = 2 m/s^2 and a_car = 4 m/s^2.
Substituting the initial positions, we have:
x_car(t) = x_bus(0) + 200 + v_car(0)t + (1/2)a_car*t^2
x_bus(t) = x_bus(0) + (1/2)a_bus*t^2
Since the car catches up with the bus, their positions will be equal at the time of catching up. Therefore, we can set x_car(t) = x_bus(t):
x_bus(0) + (1/2)a_bus*t^2 = x_bus(0) + 200 + v_car(0)t + (1/2)a_car*t^2
Simplifying the equation, we have:
(1/2)a_bus*t^2 - (1/2)a_car*t^2 - v_car(0)t - 200 = 0
Since we are looking for the time at which the car catches up with the bus, we can solve this quadratic equation to find t. However, we need more information about the initial velocity of the car, v_car(0), in order to solve the equation.
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A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer?
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A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer?.
A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2 and the car has acceleration 4 m/s2. The car will catch up with the bus after a time of:a)√110 sb)√120 sc)10 √2 sd)15 sCorrect answer is option 'D'. Can you explain this answer?.
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