SUVAT question help

The deceleration is what's confusing me tbh

Deceleration is just negative acceleration, so for the last part of the journey, the acceleration is -2 ms-1.

I would suggest sketching a velocity-time graph - it should make this quite easy.

I'm struggling with a SUVAT question I received as hw, any help would be appreciated

A train travels along a straight piece of track between 2 stations A and B. The train starts from rest at A and accelerates at 1.25m/s^2 until it reaches a speed of 20m/s. It then travels at this speed for a distance of 1560m and then decelerates at 2m/s^2 to come to rest at B.

A) find the distance from A to B

B) find the total time take for the whole journey

C) find the average speed of the journey

The deceleration is what's confusing me tbh

Firstly always draw out a diagram and state the SUVATS, you can do this question in parts:


Part 1, finding S when accelerating
S = ?
U = 0 (Starts from rest)
V = 20
A = 1.25
T = ?

V^2 = u^2 + 2as
20^2 = 0 + 2(1.25)S
400 = 2.5 S
S = 160 m

(So it took 160m to reach 20m/s at that acceleration and then it travels for another 1560m at constant speed. So it does a total of 1720m before decelerating)

Part 2, finding a distance it covered whilst decelerating. That's negative acceleration. So:

S = ?
U = 20
V = 0 (comes to rest)
A = -2
T = ?

V^2 = U^2 + 2as
0 = 20^2 + 2(-2)s
0 = 400 -4S
-400 = -4S
S = 100 m

Add this to the 1720m it already covered before decelerating and you get 1820m

So a) 1820m

Part 3, find the time taken when the train is accelerating
So this time you are looking for T, but don't use the answer you got in the previous question unless you need to, in case it might be wrong

S = ?
U = 0
V = 20
A = 1.25
T = ?

V = U + at
20 = 0 + (1.25)t
20 = 1.25t
T = 16 seconds

Part 4, find the time it stayed at constant speed

S = 1560 m
U = 20
V = ?
A = 0
T =?

S = ut + 1/2at^2
1560 = 20t + 1/2(0)t^2
1560 = 20t
T = 78 seconds

Part 5, see how long it spent decelerating,

S = ?
U = 20
V = 0 (comes to rest)
A = -2
T = ?

v = u + at
0 = 20 -2t
-20 = -2t
T = 10 seconds

16 + 78 + 10 = 104 seconds
So b) 104 seconds

Part 6, find the average speed

Speed = Distance/Time

So Speed = 1820/104 = 13m/s

Meaning c) 13m/s

Hope this helps