M1 help

I keep getting the wrong answer for this question

A train, travelling on a straight track, is slowing down with constant deceleration 0.6ms-2. The train passes one signal with speed 72km h-1 and a second signal 25s later. Find a) the speed, in km h-1, of the train as it passes the second signal, b)the distance between the signals.

For a) I used v=u+at but I got the wrong answer

Any help?

Reply 1

username219321

I suspect you didn't convert the units - the decleration is in m/s, the speed is in km/h.

Reply 2

Mist786

matt2k8

I suspect you didn't convert the units - the decleration is in m/s, the speed is in km/h.

Stupid question but how do you convert the units?

Reply 3

hassan_s

s=?
u=72*1000=72000/3600=20
v=?
a=-0.6
t=25

a)v=u+at
v=20+(-0.6*25)
v=20-15
v=5ms-1

b)s=((u+v)/2)t
s=((20+5)/2)25
s=12.5*25
s=312.5m

Reply 4

Mist786

hassan_s

s=?
u=72*1000=72000/3600=20
v=?
a=-0.6
t=25

a)v=u+at
v=20+(-0.6*25)
v=20-15
v=5ms-1

b)s=((u+v)/2)t
s=((20+5)/2)25
s=12.5*25
s=312.5m

The answer for is a) is 18 according to the answers at the back of the book

Reply 5

hassan_s

Mist786

The answer for is a) is 18 according to the answers at the back of the book

Really? It does work out at 5.

It is because, to convert 72kmh to m/s, you multiply it by 1000 (72000) to get m/h, then divide by 3600 to get m/s. This is 20. Then, you multiply a (-0.6) by t(25), which gives the answer 15. Then, according to v=u+at, you just do 20-15 to get 5.

Reply 6

Mist786

hassan_s

Its okay

I emailed my tutor, I think it was a mis-print

anyways thanks

Reply 7

mathsisverycool

hello, only 9 years later do i stumble upon the question i was stuck on, in this chat. 5ms-1 is 18kmh-1, thats why the answer is 18. lol its not a misprint :smile: