MHB Maths Mechanics (M1): Calculate Initial Velocity & Acceleration

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The discussion revolves around calculating the initial velocity and acceleration of a particle moving with constant acceleration. The user set up two equations based on the particle's positions at specific times using the suvat equations but encountered discrepancies with the answer book. After reviewing the equations, it was determined that the correct acceleration is -2/5 m/s² and the initial velocity is 15 m/s. The user realized their method was correct, but they had made an arithmetic error in their calculations. This highlights the importance of careful arithmetic in solving physics problems.
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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!
 
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Needhelp said:
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
 
Ahh so my method was right, it was just a stupid arithmetic error somewhere! Thank you so much!
 
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