two friends started from same point with some constant acceleration of 5m/s second one starts journey 5 sec later. find the time taken by 1 St one such that he is 100m ahead of 2 nd one
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Answers
Given that, Two friends started from same point with some constant acceleration of 5m/s,The second friend started 5 seconds later.
We need to find the time taken by the first first such that he is 100 m ahead of the second one.
First One :-
- Initial velocity, u₁ = 0 m/s [ starts from rest ]
- Constant Acceleration, a = 5 m/s
- Starting time, t₁ = 0 s
Second One :-
- Initial velocity, u₂ = 0 m/s
- Constant Acceleration, a = 5 m/s
- Starting time, t₂ = 5 s
Let the time taken by the second friend be t
Then the time taken by the first friend would be t + 5
So, Using second Equation of motion, we have
⇒ s = ut + 1/2at²
⇒ s = 1/2 × 5 × t²
⇒ s = 5t² / 2 ...(1)
Similarly, Displacement made by the second one
⇒ s₂ = ut + 1/2 at²
⇒ s₂ = 1/2 × 5 × (t + 5)²
⇒ s₂ = 5(t + 5)² / 2 ...(2)
Now, that we have to find the time taken by the first friend such that he is 100 m ahead:
∴ s - s₂ = 5(t + 5)² / 2 - 5t² / 2
⇒ 100 = (5t² + 125 + 50t - 5t²) / 2
⇒ 200 = 125 + 50t
⇒ 75 = 50t
⇒ t = 75/50
⇒ t = 1.5 s
So,
⇒ Time taken by the first friend = t + 5
⇒ T = 1.5 + 5
⇒ T = 6.5
So, Time taken by the first friend would be 6.5 s.