# Physics Trick Question

1. Oct 2, 2011

### Maty

OK, so I had this physics homework question about 2 weeks ago, and I thought I would post it here to see if anyone can figure it out. Note: this homework assignment is overdue and can't go back, I am just asking to see if anyone can get it. Reason I am asking is because I asked 2 physics teachers in my high school and neither of them figured it out... I just found out about this website and a I am assuming everyone here is smart so I'm wondering if anyone here can get it.

A cab driver picks up a customer and delivers her 3.10 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration five times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases.

p.s. I don't know what the answer is and neither can I check. It's an online assignment and so I can't change any answers from assignments that are overdue. And I guess since this isn't a homework help problem, more like see if you can figure it out for the heck of it, I post it here.

2. Oct 2, 2011

### Staff: Mentor

Seems fairly straightforward to me. If you need a hint, start by comparing the time of the acceleration phase to the time of the deceleration phase.

3. Oct 2, 2011

### PAllen

I'll give one more hint. If you were asked to find the time of turnaround or the magnitude of acceleration, the problem would be under-determined (as it superficially looks). However, you really only need the ratio of distance (combined with the total given distance). In this case it is well determined, and solvable with the straightforward manipulation of the basic acceleration / distance /velocity formulas.

4. Oct 2, 2011

### PeterO

Since the cab was always accelerating [or decelerating] the average speed will be half the maximum speed. maximum speed was the speed limit. What is the speed limit there. In the suburbs here maximum speed is 60 km/h - so average speed is 30km/h.
For simplicity here I will claim the maximum speed limit is 62 km/h, so the average speed is 31 km/hr.

That means 1/10 of an hour [6 minutes] to cover the 3.1 km.

Deceleration was at 5 times the rate, so only takes 1/5 the time

Accelerate for 5 minutes, decelerate for 1 minute. Since average speed for each phase is 31 km/h you can work out the distances.

ALSO: you don't need to know the speed limit - those distances and times apply regardless.

5. Oct 2, 2011

### PAllen

Well, except the trick is that the speed limit isn't given and therefore you cannot determine acceleration times or magnitude of acceleration, but you can still determine the distances. Solving it for a speed limit you choose is not an adequate solution without justifying that the speed limit doesn't matter - in which case why bother assuming one.

Last edited: Oct 2, 2011
6. Oct 2, 2011

### PAllen

The times do not apply. Only the ratio of times apply.

7. Oct 2, 2011

### DaveC426913

You could draw this on a graph. Set the decel slope to be -5x the accel slope.

8. Oct 2, 2011

### PAllen

Well, in the spirit of the problem, I think you don't need to pick particular values of anything you don't know (slope, speed limit etc.). In fact, a give away hint is without assuming even the ratio of 5, just the set up of the problem, you should aim to derive that:

x + x * (t2/t1) = D

where x is the distance before deceleration begins, t1 is time until deceleration begins, t2 is deceleration time, D is total distance.