Mechanics question

Can anyone please help with this question. Two motorcycles P and Q move in the same direction along a straight horizontal road. At t=0 P sets off with speed 1m/s from the point O and moves with constant acceleration 2m/s2. Four seconds later Q sets off in pursuit from O with initial speed 16m/s and moves with constant acceleration 1m/s2. Find a) the times between which Q is in front of P and b) the distance from O at which P overtakes Q? Thanks.

What have you tried?

OP

Original post by old_engineer

What have you tried?

I tried to draw a speed time graph but as I don't know what time they end I couldn't use it. And a distance time graph doesn't take into account the initial speeds. I'm guessing I need to use suvat? But I don't know which equation.

Reply 3

old_engineer

11

The displacement of P can be represented by s = ut + 0.5at^2 as you know the starting speed and the acceleration. You can also do the same for Q, except you must replace t by (t-4) to take account of Q setting off at t = 4. Hint: that also means t^2 must be replaced by (t-4)^2 in the SUVAT formula.

To find the time(s) at which P and Q have the same displacements you set the two s values as equal, rearrange, and solve the quadratic.

Reply 4

ElizaRose

OP

Original post by old_engineer

The displacement of P can be represented by s = ut + 0.5at^2 as you know the starting speed and the acceleration. You can also do the same for Q, except you must replace t by (t-4) to take account of Q setting off at t = 4. Hint: that also means t^2 must be replaced by (t-4)^2 in the SUVAT formula.

To find the time(s) at which P and Q have the same displacements you set the two s values as equal, rearrange, and solve the quadratic.

Thankyou so much! That helps a lot.

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