# How do you find: 1) the acceleration 2) the distance travelled at maximum speed of this word problem?

A train takes 80 seconds to travel between two stations 1639m apart.

Starting from rest at one station, it is uniformly accelerating to a speed of 22m/s

and maintains this speed until it is brought to rest at the second station by

constant retardation. The time taken to accelerate is twice the time taken to

decelerate.

Draw a velocity-time graph to illustrate the journey and, hence, find:

i) the acceleration

ii) the distance travelled at maximum speed

### 3 Answers

- Anonymous1 month agoFavourite answer
Can't draw graphs here. But here's the calculation.

We are told “The time taken to accelerate is twice the time taken todecelerate."

Call time to decelerate T.

Time to accelerate = 2T

Time at max. speed = 80 – T – 2T = 80-3T

Average speed accelerating = (vi + vf)/2 = (0+22)/2 = 11m/s

Average velocity decelerating = (vi + vf)/2 = (22+0)/2 = 11m/s

Distance accelerating = avge.vel.* time = 11*2T = 22T

Distance decelerating = avge.vel.* time = 11*T = 11T

Distance at max. speed = vel.* time = 22*(80-3T) = 1760 – 66T

Total distance = 1639m so:

22T + 11T + 1760 – 66 T = 1639

33T = 121

T = 11/3 s

Acceleration = Δv/Δt = (22-0)/(2*11/3) = 3 m/s²

Distance at max. speed = – 66*(11/3) = 1518m

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- Anonymous1 month ago
1639 = V*t/2+V*t/4+V(80-1.5t)

1639 = 22*t*(1/2+1/4)+22*80-33t

(22*80-1639) = t(33-22*3/4)

t = (22*80-1639) / (33-22*3/4) = 22/3 of sec

acceleraz. a = V/t = 22*3/22 = 3.00 m/sec^2

deceleraz. a' = -2a = -6.00 m/sec^2

time t' at constant speed = 80-(22/3+22/6) = 80-66/6 = 80-11 = 69.0 sec

space @ constant speed Scs = t'*V = 69*22 = 1518 m

space @ cariable speed Svs = 22*t*3/4 = 22*22/3*3/4 = 22^2/4 = 121 m

total space S = Scs+Svs = 1518+121 = 1639 m

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- MarkLv 71 month ago
I invite you to do your own homework. I certainly don't want to go over abridge YOU "designed", ever... (Oh, I am a civil engineer.)

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