Simultaneously going wrong

[KevinE]

Ok so I want to post the question first and then what I've done so far.

So 2 cars traveling at different speeds. The first car is traveling with an initial speed of
5m/s = U and an acceleration of 4m/s^2

Second car
Initial speed 8m/s and a deceleration of -3m/s^2

Use the formula - v^2 = U^2 + 2as. Calculate the distance (s) both cars have traveled. When they are travelling at the same speed V

 

[KevinE]

Hang on don't bother, I've noticed something I've done wrong. bbl... maybe...

 

[KevinE]

Sorry wasn't a mistake

Ok

v^2 = u^2 + 2as

v^2 = 5^2 + 2 * 4s
v^2 = 8^2 + 2 * -3s

v^2 = 25 + 8s
v^2 = 64 + -6s

So

v^2 = 25 + 8s
v^2 = 64 - 6s ( *1.3)

_________________________________________

v^2 = 25 + 8s
1.3v^2 = 83.2 - 8s

1.3v^2 = 108.2

108.2/1.3 = 83.23

83.23 = v^2

Sqrt 83.23 = 9.12

v= 9.12



I then used v popped it back into the first equation to calculate S = 9.695

 

[KevinE]

The answer is supposed to be 13m - Where has it all gone wrong?

 

[Ishuda]

The answer is supposed to be 13m - Where has it all gone wrong?

Assuming "Use the formula - v^2 = U^2 + 2as. Calculate the distance (s) both cars have traveled. When they are travelling at the same speed V" means "Use the formula - v^2 = U^2 + 2as. Calculate the distance (s) both cars have traveled when they are travelling at the same speed V", the two s's are not the same. We have
v₁² = 25 + 8 s₁
v₂² = 64 - 6 s₂
and, according to the problem statement, v₁=v₂.

 

[KevinE]

Hmmm I think I see so it should have been 2.3v^2?

 

[Ishuda]

The answer is supposed to be 13m - Where has it all gone wrong?

Hi KevinE

I'm confused. If the answer is the single value 13m then that is a distance but what distance is it? Is it the sum of the distances the cars have traveled? Or just what is the 13m supposed to be an answer to?

Working the problem, I don't get that answer. Ignoring the 'use v² = ...' for a moment
we have these equations
s(t) = \(\displaystyle \int_{t_0}^t v(x) dx\) + s(t₀)
v(t)= \(\displaystyle \int_{t_0}^t a(x) dx\) + v(t₀)
Assuming a constant acceleration a as given we then have
a(t) = a
v(t) = a (t - t₀) + v(t₀)
s(t) = 0.5 a (t - t₀)² + v(t₀) (t-t₀) + s(t₀)
Note that
v²(t) = v²(t₀) + 2 a s(t)
if s(t₀)=0

Setting the starting time t₀ to 0 and assuming an s(0) of zero for both cars, we have
v₁(t) = 4 t + 5
s₁(t) = 2 t² + 5 t
v₂(t) = -3 t + 8
s₂(t) = -1.5 t² + 8 t
where the subscripts 1 and 2 indicate the first and second car. If the speeds are equal at a time T then v₁(T) = v₂(T) or
4 T + 5 = -3 T + 8
which results in
T = 3/7
v₁(T) = v₂(T) = 47/7 m/s
s₁(T) = 2 T² + 5 T = 123/49 m
s₂(T) = -1.5 T² + 8 T = 154.5/49 m
Nowhere do I see a 13 m, not even in the sum of the two distances.

EDIT: Fix stupid mistake in general equation.