# PhysicsMaths Mechanics (M1)

#### Needhelp

##### New member
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 , 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
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 , 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

• Jameson

#### Needhelp

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