Constant acceleration help

A particle P passes through point A with velocity 2.8ms^-1 and constant acceleration 0.12ms^-2. Three seconds later a second particle Q passes through A with velocity 2.4ms^-1 and constant acceleration 0.2ms^-2.

Write down expressions for the displacements of P and Q from A, in terms of t, where t s is the time after P passed through A.

Im confused on where to start with this problem, do I have to use SUVAT equations?

Reply 1

RogerOxon

21

Yes.

s=ut+\frac{at^2}{2}

Reply 2

charlesbrowns

OP

5

Original post by RogerOxon

Yes.

s=ut+\frac{at^2}{2}

Ok, so I got
For P:
S = 2.8(t) + (0.12)t² / 2 , which simplifies to S = 2.8t + 0.06t².

For Q:
S = 2.4(t) + (0.2)t² /2 , which simplifies to S = 2.4 + 0.1t².

Is this correct so far?

Reply 3

RogerOxon

21

Original post by charlesbrowns

Ok, so I got
For P:
S = 2.8(t) + (0.12)t² / 2 , which simplifies to S = 2.8t + 0.06t².

For Q:
S = 2.4(t) + (0.2)t² /2 , which simplifies to S = 2.4 + 0.1t².

Is this correct so far?

Check your equation for Q. You need to use the same t as for P, but Q starts 3s later.

Reply 4

RogerOxon

21

Q passes A at t=3s. You therefore need to replace 't' in your equation for Q with an expression that equals 0 at t=3s. Can you see what that is?

Reply 5

charlesbrowns

OP

5

Original post by RogerOxon

Q passes A at t=3s. You therefore need to replace 't' in your equation for Q with an expression that equals 0 at t=3s. Can you see what that is?

Im sorry but I do not see how I can do this, could you possibly post the solution and therefore I can work backwards to see where it is that I am getting confused.

Reply 6

RogerOxon

21

Original post by charlesbrowns

Im sorry but I do not see how I can do this, could you possibly post the solution and therefore I can work backwards to see where it is that I am getting confused.

Your expresion for 't' in the equation for Q needs to be 0 when t=3s. It needs to be 1 when t=4s, and 2 when t=5s. As Q starts at t=3s, but the equation assumes that it starts at t=0s.

If I were to write (t-1) instead of t, then that expression would equal 0 at t=1s.
If I were to write (t-2) instead of t, then that expression would equal 0 at t=2s.

What do I need to replace t with for it to equal 0 at t=3s?

Reply 7

charlesbrowns

OP

5

Original post by RogerOxon

Your expresion for 't' in the equation for Q needs to be 0 when t=3s. It needs to be 1 when t=4s, and 2 when t=5s. As Q starts at t=3s, but the equation assumes that it starts at t=0s.

If I were to write (t-1) instead of t, then that expression would equal 0 at t=1s.
If I were to write (t-2) instead of t, then that expression would equal 0 at t=2s.

What do I need to replace t with for it to equal 0 at t=3s?

Wow, I feel stupid. I dont know why I couldnt see that.

So now For Q:
S = 2.4(t-3)+0.2(t-3)² / 2, which simplifies to 0.1t² + 1.8t -6.3

and for part b, it says show that when particles meet, 2t²-50t-315=0

I would equate my equation for P and Q:
2.8t + 0.06t² = 0.1t² + 1.8t -6.3
simplify to get:
0.04t²-1t-6.3 = 0
and show that by multiplying by 50, I get 2t²-50t-315=0.
Right?

Thank you for the help.

(edited 6 years ago)

Looks right to me.

OP

Original post by RogerOxon

Looks right to me.

Now it says, find the distance from A when the two particles meet.

Am I supposed to find t from the equaton when the two particles meet and sub that into another SUVAT equation?

Reply 10

RogerOxon

21

Original post by charlesbrowns

Now it says, find the distance from A when the two particles meet.

Am I supposed to find t from the equaton when the two particles meet and sub that into another SUVAT equation?

Yes. Take the positive solution for t and use it in either of your equations - they'll give the same value for s.

Constant acceleration help

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