[Guest]

My mechanical engineering lecturer set this question for us a few weeks ago and I thought is was quite good.

A coach starting from rest uniformly accelerates during the first 1 km, travels 2.5km at its maximum and is brought to rest in 0.5km. If the time for the whole journey is 8 mins, find the uniform acceleration at the start.

You may need to draw a VT graph

[Guest]

it was quite easy question.....i wonder if it was really given by a mechanical engineering lecturer

accelaration = 8.5078125 km/h

[curr3nt]

it was quite easy question.....i wonder if it was really given by a mechanical engineering lecturer

accelaration = 8.5078125 km/h

Did you mean 8.5078125 km/m²?

Using |v|² = |u|² + 2a * s

|v|² = 2(8.5078125 km/m²) * 1 km

|v|² = 17.015625 km²/m²

v = 4.125 km/m

This makes the time spent accelerating 0.4848 minutes using v = u + at.

At 4.125 km/m it would take 1.65 minutes to go the 2.5 km.

Are you trying to say it took about 6 minutes to decelerate? Did I miss something?

---

Are we to assume a uniform deceleration?

Edited November 30, 2011 by curr3nt

[CaptainEd]

I'm with curr3nt--it appears we don't know enough for a unique answer. Since it takes less time to decelerate than it did to accelerate, it appears we have to discover at least two values, the uniform acceleration and the separate uniform (or otherwise) deceleration. It feels like we have too much latitude (or, another way of stating it, too few constraints.

[CaptainEd]

If we assume a (different) uniform deceleration as well, it is pretty heavily constrained. My first attempts at initial accelerations were way too high--if we decelerated uniformly, we traveled more than the allotted distance.

a = .23635 km/min/min

t(acceleration) = 2.91 min

t(freewheeling) = 3.63 min

t(deceleration) = 1.45 min

Sorry not to have a closed form.

Edited November 30, 2011 by CaptainEd

[CaptainEd]

The closed form, with derivation, clumsy and wordy though it be...

Let's speak of the three movement phases as phases 1 (acceleration), 2 (freewheeling), 3 (deceleration)

First, assume a constant deceleration different from the constant acceleration.

a1 is acceleration

a2 is zero, at velocity v

a3 is deceleration

v = a1 * t1

and

v = a3 * t3

Look at the constraints due to distance traveled (s)

s1 = 1 km = 1/2 a1 * t1*t1

s3 = .5 km = 1/2 a3 * t3*t3

Since a1 * t1 = a3 * t3, we can rewrite these as

s1 = 1 = 1/2 a1*t1 * t1

s3 = .5 = 1/2 a1*t1 * t3

2*s3 = 1 = a1*t1 * t3

So 1/2 a1*t1 * t1 = a1*t1 * t3

t1/2 = t3

In other words, in order to decelerate in half the distance, we must decelerate in half the time

--------------------

Now, express the times for the phases in terms of the initial acceleration.

In phase 1, s = .5 * a1 * t1 * t1 = 1 km

t1 = sqrt(2/a1) = 2/sqrt(2*a1)

This results in velocity v = a1 * t1 = sqrt(2*a1)

In phase 2, s2 = v * t2 = 2.5 km

t2 = 2.5/sqrt(2*a1)

In phase 3, t3 = t1/2

t3 = 1/sqrt(2*a1)

so, since 8 min = t1 + t2 + t3 = 5.5/sqrt(2*a1),

sqrt(2*a1) = 5.5/8

2*a1 = (5.5/8)^2

a1 = .5*(5.5/8)^2 = .236328 km/min/min

Thank you for the interesting problem--it seemed like we didn't know enough, but with the right sort of assumption, it became tractable.

Edited November 30, 2011 by CaptainEd

[CaptainEd]

Converting from km/min/min to km/hr/hr, I get acceleration of 850.7813, which is 100 times Arjit's answer. Does one of us have a decimal point in the wrong place?

Edited November 30, 2011 by CaptainEd

[curr3nt]

With a₁ = 121/512 km/m²

v = 11/16 km/m

t₁ = 32/11 m

t₂ = 40/11 m

t₃ = 16/11 m

t₁ + t₂ + t₃ = 88/11 m = 8 m

Looks like it works.

[Guest]

I'm with curr3nt--it appears we don't know enough for a unique answer. Since it takes less time to decelerate than it did to accelerate, it appears we have to discover at least two values, the uniform acceleration and the separate uniform (or otherwise) deceleration. It feels like we have too much latitude (or, another way of stating it, too few constraints.

It's true that in the question we don't have enough information to solve it straight away but it is possible to calculate the maximum velocity and then work out the acceleration.

Total time = 480s

s1 = 1000m

s2 = 2500m

s3 = 500m

2*s1+2*s3+s2 = 5500

5500/480 = 11.4583m/s

[Guest]

But the issue is with the deceleration at the end.

We are given a time quantifier, but if deceleration isn't uniform then the answer can change wildly.

There is a vast difference between slamming on the brakes at the end, gradual (uniform) slowing down, and hitting the brakes at the start then rolling to the end.

Possibility

1) = Short time

2) = Can be calculated

3) = Long deceleration time