Question

Cyclist A, who has constant speed 4.0 m s−1, overtakes cyclist B, who is stationary, at time t=0.

2.0 seconds later, cyclist B begins to accelerate uniformly at 1.0 m s−2.

Calculate the time at which cyclist B overtakes cyclist A.

Give your answer to an appropriate number of significant figures.

Answer 1

Answer:

See the above solution.

Thanks.

Answer 2

The answer is 10 seconds.

The answer is 10 seconds. GIVEN

Speed of cyclist A = 4.0 m/s

Acceleration of Cyclist B = 10 m/s²

TO FIND

the time at which cyclist B overtakes cyclist A.

SOLUTION

We can simply solve the above problem as follows;

Speed of Cyclist A, u₁ = 4 m/s

Acceleration of A, a₁ = 0 (Constant speed)

Speed of Cyclist B, u₂ = 0 m/s

Acceleration of Cyclist B, after 2 seconds a₂ = 1 m/s²

Relative acceleration of A, a₁₂ = a₁ - a₂ = -1 m/s²

Relative speed of A, u₁₂ = u₁ - u₂ = 4 m/s

Time after wich the second cyclist starts accelerating, T= t + 2 sec

Relative displacement,

S₁₂ = u₁₂T + (1/2)a₁₂T²

0 = 4T + T²/2

4T = T²/2

T = 8 seconds

Total time taken = 2 + 8 = 10 seconds.

Hence, The answer is 10 seconds.

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