8.Two trams , tram A and tram B, run on parallel straight horizontal tracks . Initially the two trams are at rest in the depot and level with each other. At time t=0 , tram A starts to move. Tram A moves with constant acceleration 2ms^-2 for 5 seconds and then continues to move along the track at constant speed. At time t=20 seconds, tramB starts from rest and moves in the same direction as tram A. Tram B moves with constant acceleration 3ms^-2 for 4 seconds and then continues to move along the track at constant speed. The trams are modelled as particles. (a) Sketch,on the same axes,a speed -time graph for the motion of tramA and a speed-time graph for the motion of tramB from t=0 to the instant when tramB overtakes tram A. (3) At the instant when the two trams are moving with the same speed, tramA is d metres in front of tramB (b) Find the value of d. (c) Find the distance of the trams from the depot at the instant when tramB overtake tram A (5)

Let's correct and refine the calculations step-by-step.<br /><br />### (a) Speed-Time Graphs<br /><br />1. **Tram A:**<br /> - Accelerates at \(2 \, \text{m/s}^2\) for 5 seconds.<br /> - Speed at \(t = 5 \, \text{s}\): \(v_A = 2 \times 5 = 10 \, \text{m/s}\).<br /> - Moves at constant speed of \(10 \, \text{m/s}\) after 5 seconds.<br /><br />2. **Tram B:**<br /> - Accelerates at \(3 \, \text{m/s}^2\) for 4 seconds.<br /> - Speed at \(t = 4 \, \text{s}\): \(v_B = 3 \times 4 = 12 \, \text{m/s}\).<br /> - Moves at constant speed of \(12 \, \text{m/s}\) after 4 seconds.<br /><br />### (b) Distance Calculation<br /><br />To find the distance \(d\) when the trams are moving at the same speed:<br /><br />- Tram A reaches \(10 \, \text{m/s}\) at \(t = 5 \, \text{s}\).<br />- Tram B reaches \(12 \, \text{m/s}\) at \(t = 4 \, \text{s}\), but it starts 20 seconds later.<br /><br />At \(t = 20 \, \text{s}\), Tram A's speed is \(10 \, \text{m/s}\) and Tram B's speed is \(12 \, \text{m/s}\).<br /><br />Distance traveled by Tram A in 20 seconds:<br />\[ s_A = \frac{1}{2} \times 2 \times 5^2 + 10 \times (20 - 5) \]<br />\[ s_A = 25 + 150 = 175 \, \text{m} \]<br /><br />Distance traveled by Tram B in 4 seconds:<br />\[ s_B = \frac{1}{2} \times 3 \times 4^2 = 24 \, \text{m} \]<br /><br />Distance traveled by Tram B in the remaining time (16 seconds):<br />\[ s_B' = 12 \times 16 = 192 \, \text{m} \]<br /><br />Total distance traveled by Tram B:<br />\[ s_B_{\text{total}} = 24 + 192 = 216 \, \text{m} \]<br /><br />Distance \(d\) when they are at the same speed:<br />\[ d = s_A - s_B_{\text{total}} \]<br />\[ d = 175 - 216 = -41 \, \text{m} \]<br /><br />This indicates that Tram B overtakes Tram A before \(t = 20 \, \text{s}\). We need to find the exact time \(t\) when their speeds are equal.<br /><br />### (c) Finding the Overtake Time<br /><br />Set the speeds equal:<br />\[ 10 = 2(t - 5) \]<br />\[ 10 = 2t - 10 \]<br />\[ 20 = 2t \]<br />\[ t = 10 \, \text{s} \]<br /><br />At \(t = 10 \, \text{s}\), Tram B has been moving for \(10 \, \text{s}\) (since it started at \(t = 20 \, \text{s}\)).<br /><br />Distance traveled by Tram B:<br />\[ s_B = \frac{1}{2} \times 3 \times 10^2 = 150 \, \text{m} \]<br /><br />Distance traveled by Tram A:<br />\[ s_A = 25 + 10 \times (10 - 5) = 25 + 50 = 75 \, \text{m} \]<br /><br />Distance \(d\) when Tram B overtakes Tram A:<br />\[ d = 150 - 75 = 75 \, \text{m} \]<br /><br />### Summary<br /><br />(a) The speed-time graphs for Tram A and Tram B will show Tram A accelerating to \(10 \, \text{m/s}\) in 5 seconds and then moving at constant speed, while Tram B accelerates to \(12 \, \text{m/s}\) in 4 seconds and then moves at constant speed.<br /><br />(b) The distance \(d\) when Tram A is in front of Tram B is \(75 \, \text{m}\).<br /><br />(c) The distance of the trams from the depot when Tram B overtakes Tram A is \(150 \, \text{m}\) for Tram B and \(75 \, \text{m}\) for Tram A.