Question

the train accelerates from 30 km/h to 45 km/h in 15 secs. a. find its acceleration. b. distance it travels during this time

Answer 1

we use the equations of motion to solve this. We know the initial speed \(u=30\) the final speed \(v=45\) and we know time \(t=15\). The acceleration \(a\) can be found from \[v=u+at\] rearranging we get \[a=\frac{v-u}{t}=\frac{45-30}{15}=1\]so acceleration is 1 \(km/hour^2\)

The distance s can be found by using this acceleration with either the equation \[s=ut+\frac{1}{2}at^2\]or\[v^2=u^2+2as\]using the first equation we have \[s=(30)(15)+\frac{1}{2}(1)(15)^2=562.5\]using the second equation (after rearranging) we get \[a=\frac{v^2-u^2}{2}=\frac{(45)^2-(30)^2}{2}=562.5\]So the answer is 562.5 km.

Answer 2

i think we got different answers because you didn't convert 15 seconds to hours? :o

Answer 3

Sorry I misread the question. I will redo the post correcting it

we use the equations of motion to solve this. We know the initial speed \(u=30 \rm{km/h}=8.333\rm{m/s}\) the final speed \(v=45\rm{km/h}=12.5 \rm{m/s}\) and we know time \(t=15\). The acceleration a can be found from
\[v=u+at\] rearranging we get
\[a=\frac{v−u}{t}=\frac{12.5−8.333}{15}=0.2778\]
so acceleration is 0.2778 mm/s\(^2\)2. The distance \(s\) can be found by using this acceleration with either the equation
\[s=ut+\frac{1}{2}at^2\]
or

\[v^2=u^2+2as\]
using the first equation we have
\[s=(8.333)(15)+\frac{1}{2}(0.2778)(15)^2=156.2475\]
using the second equation (after rearranging) we get
\[s=\frac{v^2−u^2}{2a}=\frac{(12.5)^2−(8.333)^2}{2(0.2778)}=156.2475\]
So the answer is 156.25 m.

Apologies for any confusion caused in my error.

Answer 4

Just spotted a typo. The line "so acceleration is 0.2778 mm/s22. The distance s can be found by using this acceleration with either the equation" should read "so acceleration is 0.2778 m/s\(^2\). The distance s can be found by using this acceleration with either the equation" it most certainly is not millimetres.