Applied Maths question re: linear motion.

oq4v3ht0u76kf2

Acceleration: a
Deceleration: b
Distance: s
Starts and ends at rest (0).

Show that if there is a speed limit of 'v' m/s the time taken to complete the journey will be:

v/2a + v/2b + s/v

I am finding this question very hard!

Okay, we have three seperate phases:

1. Acceleration
v = u + at => t = v/a


2. Cruising
s = v*t => t = s/v (I've gotten this bit right.)

3. Deceleration
v = u + at => 0 = v + bt => t = -v/b
[v = 0, ends at rest; u = v]

Can anyone point me in the right direction?

[Exercise 2C, Question 20, Part (i) in Fundamental Applied Mathematics in case anyone is interested.]

DeVore

How come the deceleration period doesnt mention "b". I think thats your first mistake. Also I am suspect about zeroing v in phase 3 too but I need to reread some linear mechanics before I go telling you that.

Pretty sure about the "b" thing though

DeV.

DeVore

Right, if "v" is the speed limit then "u" in the deceleration equation can be set to "v" (since its topped out at the speed limit... no boy-racers here!)

Sorry that I cant prove the theorem for you but this is your homework I'm just being helpful

DeV.

oq4v3ht0u76kf2

It's actually revision for ye olde summer exams, not quite the leaving cert but still important (to me, anyway)... thanks for your help on the u = v thing but the a/b thing in the deceleration equation was just a slip. It's fixed now.

Sev

I dont really think you can 'point somebody in the right direction' for a question like this, I'd say everybody does it in some different way. I always hated these accelerated linear motion, I think the approach to answering these questions is so counter-intiuitive, well counter-my intuition anyway, and they never work out for me.

I assumed three phases like yourself, with three separate times t1,t2,t3 and used the fact that t1 = v/a, t3 = v/b and that t = (t1 + t2 + t3) to solve it. I may well be doing this in the most roundabout fashion.. but this is the first method I approached it with.
Anyway, ive posted my full solution to this, so if you dont want to know it.. stop reading now.

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I say....

s = s1 + s2 + s3
= (v/2)(t1) + (v)(t2) + (v/2)(t3)

add (v/2)(t1) and (v/2)(t3) to both sides, you get

s + v(t1)/2 + v(t3)/2 = v(t1 + t2 + t3)

and (t1 + t2 + t3) = t
so..

s + vt1/2 + vt3/2 = vt

divide by v on both sides

s/v + t1/2 + t3/2 = t

but.. t1 = v/a and t3 = v/b

so

t = v/2a + v/2b + s/v

article6

I always found it useful to draw a diagram, with Time on the X-axis and Velocity on the Y-axis. Then you can chart the velocity (in the form of a wedge-shape), which makes the mental work a lot easier.

dudara

A time-velocity graph is a good thing to draw here. Use the area underneath to find the distance and then make subsitutions from v = u + at for the individual times that you've assumed in order to draw the graph

the solution to the question, which is stated as part of the question, contains factors of a half, which immediately makes me think of the area of a triangle and time-velocity graphs.

thegills

Think of average speeds.

T1=V/A.
V changes from 0 to 'v' m/s with constant acceleration so V=(0+v)/2 = v/2
So T1 = (v/2)/A = v/2A

Similiarly T3 = v/2B

T2 = s/v

So Time taken = T1+T2+T3 = v/2A + s/v + v/2b

NB: You were mixing up your V and your v

Ciaran

Your mistake is that you've taken the distance travelled at constant speed as being the total distance travelled. You have to work out the distance travelled at constant speed separately, see bit in bold.

V = 0 + at1
t1 = V/a

0 = V - bt3
t3 = V/b

V^2 = 0 + 2as1
s1 = V^2/2a

0 = V^2 - 2bs3
s3 = V^2/2b

[B]s = s1 + s2 + s3
s2 = s - s1 - s3
s2 = s - V^2/2a - V^2/2b[/B]

t2 = s2/V
t2 = (s - V^2/2a - V^2/2b)/V

t = t1 + t2 + t3
  = V/a + V/b + s/V - V/2a - V/2b
  = V/2a + V/2b + s/V

M@lice

Draw your velocity vs time graph.

Then total distance traveled = Area under graph

After a quick read of the question that looks like the key right there.

Nietzschean

well one way or another, draw draw draw, you get stupidly high ammounts of marks for just diagrams and inital eqn's in applied maths, its only like 5marks for getting the solution.

that and back in my day i found lc linear motion stuff became a hell of alot easier with a diagram.....

DS

Well you can't say one way is better than the other. Different methods are better for different questions. If there's more than one change in acceleration it's much easier to use a diagram, if there's no change in acceleration then definitely equations only.