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-1 . It then travels at this speed for a distance of 1560m and then decelerates at - 2m·s-2 to come to rest at B.
Answers
Explanation:
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
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2 years ago
(Original post by Den987)
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