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Physics Maths Mechanics (M1)

Needhelp

New member
Apr 9, 2012
17
This question is probably impossibly easy, but I have been staring at it for half an hour now, and don't seem to be able to get the right answer :mad:, so any help would be great!

A particle P moves on a straight line with constant acceleration. At t=0, P passes through the point 0 on a line. When t=20, P passes through a point A, where OA=220m and when t=50, P passes through a point B where OB= 250m and AB=30m.

Calculate the initial velocity and acceleration of P.

This is what I did:

I spilt it into two equations, with my first being- 220=20u +200a and my second being- 250=50u + 1250a (using the suvat equation s= ut + 0.5at2)
I then sub in u= 11-10a as u = 220-200a into my second equation to get 250= 550-500a+1250a, and therefore my a = -3/7ms-2.

In my answer book it says the the acceleration = -0.4ms-2, and so I gave up trying to find u... does anyone know where I went wrong ( or even is the book wrong- it can be very unreliable!!)

Thank you!!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
This question is probably impossibly easy, but I have been staring at it for half an hour now, and don't seem to be able to get the right answer :mad:, so any help would be great!

A particle P moves on a straight line with constant acceleration. At t=0, P passes through the point 0 on a line. When t=20, P passes through a point A, where OA=220m and when t=50, P passes through a point B where OB= 250m and AB=30m.

Calculate the initial velocity and acceleration of P.

This is what I did:

I spilt it into two equations, with my first being- 220=20u +200a and my second being- 250=50u + 1250a (using the suvat equation s= ut + 0.5at2)
I then sub in u= 11-10a as u = 220-200a into my second equation to get 250= 550-500a+1250a, and therefore my a = -3/7ms-2.

In my answer book it says the the acceleration = -0.4ms-2, and so I gave up trying to find u... does anyone know where I went wrong ( or even is the book wrong- it can be very unreliable!!)

Thank you!!
Under constant acceleration \(a\) the position \(s\) at time \(t\) is:

\[s(t)=\frac{at^2}{2}+v_0t+s_0\]

where \(v_0\) and \(s_0\) are the velocity and position at \(t=0\).

Then you are told that:

\(s(0)=s_0=0\)

and so using this and the next thing you are told:

\(s(20)=200a+20v_0=220\)

and again:

\(s(50)=1250a+50v_0=250\)

which gives you the pair of simultaneous equations:

\(200a+20v_0=220\)
\(1250a+50v_0=250\)

to solve (multiply the first by \(5\) and the second by \(2\) and subtract), which has solution \(a=-2/5\) and \(v_0=15\)

CB
 

Needhelp

New member
Apr 9, 2012
17
Ahh so my method was right, it was just a stupid arithmetic error somewhere!! Thank you so much!