Scottish Elises

And it's got cars in it!

Two cars are travelling along the same road at different speeds. As Car One passes Car Two, Car One is doing 100 mph and Car Two is doing 70 MPH. (Car One is the faster, that's why it's number one).

At this moment a HUGE TREE falls INSTANTLY across the road and both drivers, their reactions like hair triggers SLAM their brakes on at the same exact instant.

Car Two comes to a screeching halt, it's bumper ALMOST touching the fallen tree.

How fast is Car One travelling when he hits the tree?

(Apologies if you have seen this before)

Don't know the answer,

But if Tut was driving car No 1 he would still have been accelerating hard and passed by before the tree even got near the ground.

Don't know the answer,

But if Tut was driving car No 1 he would still have been accelerating hard and passed by before the tree even got near the ground.

Depends if it had winter tyres on or not

I'll get my coat.....

0mph. He's hit three tree

Car one doesn't hit the tree - he was going in the opposite direction.

Tut would have jumoed the tree. And no, it's not a trick question!

Honestly this forum is not what it once was, once that would have been solved in seconds.

Robin has facked off

(It's 70 by the way. SEVENTY!!!)

You don't get full marks unless you show your working.......

Distance equals area under the speed time graph

I'm ashamed.
I have two Physics degrees and I couldn't get my head around that one until you explained it.

To make myself feel better I shall ask the question to my Physics-teacher wife. Unless of course she answers it correctly. Then I'll feel worse, obviously.

It caught me out at first, I tried to resolve it with time

So, I felt so embarrassed I had to prove to myself I could still do higher physics.

The only things we know are initial velocity (v) and final velocity (u) of each car.
We don't know acceleration, time or distance.
But we can assume that acceleration (a) (deceleration in this case) is the same in both cars and we know that the distance (s) to the tree is the same for both cars, even though we don't know the actual numbers.

Use formula v^2 = u^2 +2as
Rearrange;
2as = v^2 - u^2

As we know that the deceleration and distance is the same for both cars we can say that;

(v^2 - u^2) for car 1 is equal to (v^2 - u^2) for car 2
or
v(car1)^2 - u(car1)^2 = v(car2)^2 - u(car2)^2

The one we are trying to work out is u(car1) (final speed of car one) so we can rearrange;

u1^2 = (v1^2)-(v2^2)+(u2^2)

Convert 70mph into 31.3m/s and 100mph into 44.7m/s

u1^2 = 1998 - 979 + 0
u1^2 = 1018
u1 = 31.9m/s

So final speed of car 1 as it hits the tree is 31.9m/s which is 71.4mph.

Anyone still awake?