A train is timed between successive posts $A, B$ and $C$, each $2000$ m apart. It takes $100$ seconds to travel from $A$ to $B$ and $150$ seconds to travel from $B$ to $C$. The acceleration throughout the journey is uniform.
What is the acceleration?
I get $-2/75 ms^{-2}$
This, according to the website which marks it, is incorrect.
My method is:
[(Disp. $A$ to $B$/Time $A$ to $B$)-(Disp. $B$ to $C$/Time $B$ to $C$)]/total time
= [(Velocity $A$ to $B$)-(Velocity $B$ to $C$)]/total time
= Change in velocity/time
= acceleration
= $[(2000/150)-(2000/100)]/150+100 = -2/75ms^{-2}$
Can someone please explain why I'm wrong? Many thanks for your help
Best Answer
The velocity from $A$ to $B$ and from $C$ to $D$ isn’t uniform, you’re erring by taking the average acceleration instead of the exact.
Using the relation $S=ut+\frac 12 at^2$, you get two equations corresponding to the two trips in $u$ and $a$:
$$u+50a=20 \\3u + 400a=40$$ Note that the ‘$u$’ for the second trip (from $B$ to $C$) is obtained by using $v=u+at$ for the first trip (as the first trip’s final velocity is the same as the second’s initial velocity). After solving you should get$$a=-\frac{2}{25}$$