[Closed] Maths help please

Chaser has to do 5 miles in less than (4x36)seconds or 144 seconds. An hour is 3600 seconds, so faster than (3600/144)x5 = 125mph.

so the first car travels 5 miles in 5 x 36s = 180s

the second neads to cover 5 miles in 180 - 36 = 144s

144 / 5 = 28.8 seconds per mile or 125 mph

First car travels five miles. Chasing car has to travel six miles (the five plus one to catch up) in the same amount of time.

Lead car going 100, chase car has to go (6/5)*100.

120mph

Five miles from when car 1 starts or car 2 starts?

First car travels five miles. Chasing car has to travel six miles (the five plus one to catch up) in the same amount of time.

Not quite. Need to start at point first has done a mile. So second must do 5 miles in time first does 4.

In your setup, when the first has done a mile, second has caught up a bit already so doesn’t need to go so fast. In the problem, we know it’s a mile behind there.

Edit: though, rereading it - could read it your way.

so the first car travels 5 miles in 5 x 36s = 180s

the second neads to cover 5 miles in 180 – 36 = 144s

144 / 5 = 28.8 seconds per mile or 125 mph

I’m going with this

Chasing car has to travel six miles (the five plus one to catch up) in the same amount of time.

Then your calculation is for catching the lead car in six miles not the five stipulated

First car travels five miles. Chasing car has to travel six miles (the five plus one to catch up) in the same amount of time.

Lead car going 100, chase car has to go (6/5)*100.

120mph

+1

If the second car has to go the same distance in 4/5ths of the time, then it has to go 5/4ths as fast ie 125mph. Greater to catch the first car within the 5 miles.

This is Vaguebook clickbait again, isn't it.

The chaser is starting 36 seconds or 1 mile behind the 1st car.

These are different. I think you mean, chaser starts when first car has done 1 mile or, equivalent, 36s after first car. If first car has a head start of 1 mile and they start at the same time, you end up with sockpuppet’s numbers.

It does depend on the setup.

If the two cars start in the same place, one sets off at 100 and the other waits 36 seconds before setting off then it does need to do 125 to get to the five mile point at the same time.

That’s not what the OP situation sounds like.

It seems that the two cars are driving along at different speeds, and at time zero they’re one mile apart. To catch the first car when it’s done five miles the chaser needs to cover six. 120mph.

OP - can you clarify?

Edit:

The chaser is starting 36 seconds or 1 mile behind the 1st car.

These are different.

A fine point! They’re the same at 100mph, but not if the speed is otherwise.

Which car is on the conveyor belt?

300mph - will deffo catch it with 5 miles and don’t have to worry about calculating it!

Yeah, but

The chaser is starting 36 seconds or 1 mile behind the 1st car.

36 seconds equals one mile only for the car doing 100mph.  If it’s 36 seconds for one car or the other or 1 mile for one car or the other or maybe they start at different points it all becomes a bit of a rubbish question

300mph

One million mph 😀

And is car two starting from standing? How fast can it accelerate?

The chaser is starting 36 seconds or 1 mile behind the 1st car.

Too vague. As already illustrated there are two legitimate ways to interpret this sentence. Is the chaser starting a mile back with both cars at their set speeds (so it will cover 6 to the first cars 5 miles in the same time) or is the chaser waiting at the same startline and only crossing it when the first car is a mile up the road (so it will cover 5 to the first cars 4 miles in the same time)?

This is great, thanks to all replies..!!

I actually meant starting 36 seconds behind, from the same starting point as car 1. Which means that car 1 is now 1 mile in front, though the total distance to travel for both cars is still only 5 miles.

I calculated wrong but struggled to think why; I calculated that car 2 is 1 mile behind and came up with 120mph, which instinctively felt wrong but I couldn’t see why.

Again thanks to all that’s a brilliant response 👍🏻👍🏻👍🏻👍🏻

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@Wally that's brilliant (super stealth)

So the first car travels 4 miles X 36 seconds = 144 seconds. That's the time you have to work with to work out the 5 miles covered by the second car.

For what it's worth I read it that the second car was a mile behind so covering 6 miles.

The time isn't really relevant. All you need to know is that the second car has to travel 5 miles in the same time that the first car travels 4 miles at 100 mph. Its speed must be 100 x 5/4 = 125 mph.

The question is slightly ambiguous, so it could be interpreted as 6 miles and 5 miles, in which case:
100 x 6/5 = 120 mph.

are they both going in same direction?

I’m impressed the second car goes from 0 to 125mph instantly.

I’m impressed the second car goes from 0 to 125mph instantly.

You need to watch more movies. Someone escapes with a decent headstart, then the pursuers hotwire a car and catch them up in about 10 seconds. It must be because of the special gearboxes they use, where you change down a gear to go faster.

It must be because of the special gearboxes they use, where you change down a gear to go faster.

Most of the cars in movie car chases have 18speed gearboxes as well.

To catch ‘within 5 miles’ the second car would have to travel at >125mph
It is an inequality question the > sign is important

Ok, with the OP clarification it’s deffo 125.

Shows how tricky it must be to write exam papers!

On the op's diagram which provides
clarification there appears to he a substantial wall at 5 miles.

The maximum distance the back of the first car can go is 5 miles minus the length of the car after it hits a wall at 100mph.

The 2nd car just needs to keep moving at any speed and they will catch the first car within 5 miles.