At an instant, Runner 'A' runs at 6m/s with an acceleration of 2m/s^2. Runner 'B' runs at 5m/s with an acceleration of 4m/s^2, and the former (Runner A) is 2 metres ahead of the latter (runner B). How long does it take for Runner B to catch up with runner A?
Start time measurement from when the separation distance, Xa - Xb, is 2.0 m.
Xa - Xb = 2.0 +(6-5)*t + (1-2)t^2
= 2 +t -t^2 = 0
t^2 -t -2 = 0
(t-2)(t+1) = 0
Solve for t. Take the positive root.
t = 2.0 s.
Thank you! i have spent an hour on that question. Which equation was that?
s=ut+at^2/2 ?
Yes, the equation for Xa - Xb results from
s = so + uo*t + a*t^2/2
where so is the displacement at t = 0
To calculate the time it takes for Runner B to catch up with Runner A, we need to determine the distance that Runner A covers during that time.
Let's assume that it takes t seconds for Runner B to catch up with Runner A.
To calculate the distance covered by Runner A during time t, we can use the equation:
Distance = Initial Velocity * Time + (1/2) * Acceleration * Time^2
For Runner A:
Initial Velocity (u) = 6 m/s
Acceleration (a) = 2 m/s^2
Therefore, the distance covered by Runner A during time t is:
Distance_A = 6t + (1/2) * 2 * t^2
Distance_A = 6t + t^2
Now let's calculate the distance covered by Runner B during time t.
For Runner B:
Initial Velocity (u) = 5 m/s
Acceleration (a) = 4 m/s^2
The distance covered by Runner B during time t is:
Distance_B = 5t + (1/2) * 4 * t^2
Distance_B = 5t + 2t^2
Since we want Runner B to catch up with Runner A, the distance covered by Runner B should be equal to the distance covered by Runner A:
Distance_A = Distance_B
6t + t^2 = 5t + 2t^2
Simplifying the equation:
t^2 - t = 0
t(t - 1) = 0
This equation has two solutions:
1. t = 0 (we disregard this solution because it represents the initial state when both runners start)
2. t - 1 = 0
t = 1
Therefore, it takes 1 second for Runner B to catch up with Runner A.