Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

A particle is travelling in a straight line with constant acceleration.It passes through 3 different points P,Q and R, such that 2PQ=QR.The velocity of the particle at P is 5m/s and that at R is 25m/s.

the question had asked to find the ratio of time required to travel from P to Q and from Q to R.

I found it to be 1:1.

But I wonder if we can also find the acceleration, since it is constant(I may be wrong).I tried but couldn't succeed.

So my question is,

Can we find the acceleration of the particle in this case?

friendship is tan 90°.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,566

hi Niharika,

Yes it can. I made a velocity / time graph for this question. It can be done without, but I found it easier.

The straight line shows the constant acceleration. I've taken my starting distance as zero (ok as you can measure from anywhere). Let the velocity at Q be V. Also let the time from P to Q be T1 and from Q to R be T2.

The distance under a V / T graph will give the distance travelled, so we can use the area of trapeziums to say

and using v = u + at

If you use these to get T1/T2 in two different ways you can solve for V and hence get T1/T2 and a.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

Offline

**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

I found V as 15m/s.

T1/T2 is 1.

but whenever I am trying to find acceleration, it is getting cancelled out.

friendship is tan 90°.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,566

hi Niharika,

I'm sorry; I hadn't noticed that. I agree with you. I then tried this:

Choose a = 2

then T1 = T2 = 5 and PQ = 50 = 2 x QR

Now choose a = 1

then T1 = T2 = 10 and PQ = 100 = 2 x QR

etc

So it would seem that all the given properties of the question can be fixed whatever 'a' is.

So the answer to your question " Can we find the acceleration of the particle?", would appear to be no. Curious question.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

Offline

**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,062

thanks bob,

It was really curious.

friendship is tan 90°.

Offline

Pages: **1**