Can anyone help me with one of these super basic physics questions that I still don't know how to do .-. thanks!
A car travelling with a constant speed of 80 km/h passes a stationary motorcycle policeman. The policeman sets off in pursuit, accelerating uniformly to 80 km/h in 10.0 s and reaching a constant speed of 100 km/h after a further 5.0 s. At what time will the policeman catch up with the car?
To do this, the first step you need to do is construct a velocity-time graph.
You want to depict both the policeman and the car on it (remember to convert km/h to m/s, you can do this by dividing km/h by 3.6)
Your graph should show the policeman accelerating to the same speed as the car in 10seconds, and then rising above it in the next five seconds.
Next, you want to tackle the distance that the police car has moved first. You do this by calculating the area of the triangle under the graph (10 seconds) and the trapezium under the graph (from 10 to 15 seconds). The police car now continues at a constant speed, so the formula you should have is x=(distance triangle)+(distance trapezium)+(speed times t-15). The t-15 part takes into account the distance travelled over the past 15 seconds.
Next, is the distance that the car travels. Throughout all this time, it travels at a constant velocity, so you can use the formula x=vt, subbing in your value for v.
Now to find when they meet, you equate the two equations together, giving you (distance triangle)+(distance trapezium)+(speed times t-15)=vt. The only unknown variable you should have here is t, which you will find using algebra.
Let me know if you have any questions, I tried to go through each step and it got pretty long in the end.