Albert is riding his scooter at a velocity of **80km/h**, when he sees an old woman crossing the road **45m **away. He immediately steps hard on the brakes to get the maximum deceleration of **7.5m/s^2**. How far will he go before stopping?

Guest Jul 19, 2017

#1**+3 **

Best Answer

Use v^2 = u^2 + 2as

v = 0 m/s. Final velocity

u = 80000/3600 m/s. Initial velocity

a = -7.5 m/s^2. Acceleration (-ve sign indicating deceleration)

s is distance travelled in metres.

s = (80000/3600)^2/(2*7.5) m → 32.9 m

.

Alan Jul 19, 2017

#3**+1 **

Can I ask why -7.5 suddenly became 7.5 in your solution?

Guest Jul 20, 2017

edited by
Guest
Jul 20, 2017

#5**+1 **

Because the "-" sign cancels when you divide 80,000/3,600=22.2222......., which becomes negative when you move it to the left side of the equation. So that you have: (22.2222...)^2 =-493.827 / (2* -7.5)=~32.92m. Or, you could cancel the negative sign by multiplying both sides by -1, which is the same thing.

Guest Jul 20, 2017

#6**0 **

*So that you have: (22.2222...)^2 =-493.827 / (2* -7.5)=~32.92m *

This is a wonder of slop to** behold! **This makes it very clear that (22.2222...)^2 = ~32.92m. I could guess who the bone brain (BB) is that presented** this lazy slop**.

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**Sir Alan assumes because you are doing grade 11 physics you should be able to do grade 8 algebra.**

To answer your question for “+” becoming “/” in the solution: It is not the same equation. Alan’s solves for “s” in the second equation.

Here’s a predigested presentation:

\(v^2=u^2+2as\\ \text {Solve for s }\\ s=(v^2-u^2)/(2a)\\ s=\dfrac{(0^2-22.22^2)}{(2(-7.5))} \leftarrow \small \text {(-7.5 because scooter is decelerating) }\\ s=\dfrac{(-(22.22^2)}{-15} \leftarrow \small \text {(Notice the negative divided by a negative. This will give a positive solution.)}\\ s=\dfrac{(-493.82)}{-15}\\ s=32.92m \\\)

Guest Jul 20, 2017

#2**+2 **

Albert is riding his scooter at a velocity of 80km/h, when he sees an old woman crossing the road 45m away. He immediately steps hard on the brakes to get the maximum deceleration of 7.5m/s^2. How far will he go before stopping?

Do you need to do this with claculus or with physics formulas?

__FORMULA METHOD__

Here are the motion formulas that you should learn.

\(80km/hour =80000/(60*60) m/sec=800/36 m/sec=800/36 m/sec=22.\dot2 m/s\\\)

u=22.2 m/s

v=0

t=?

a=-7.5

s=?

I want stopping distance, s and i do not have t so that means I can use the last formula

\(v^2=u^2+2as\\ 0=22.\dot2^2+2*-7.5*s\\ s\approx 32.92m\)

The scooter can stop in 33m so Albert should stop in time.

NOW USING CALCULUS

\( \ddot x=-7.5\\ \dot x = -7.5t+22.\dot2\\ x=\frac{-7.5t^2}{2}+22.\dot2t\\ x=-3.75t^2+22.\dot2t \)

When \(\dot x = 0\)

\(0=-7.5t+22.\dot2\\ t\approx2.963sec\\ x\approx-3.75*2.963^2+22.\dot 2*2.963\\ x\approx 32.92m \)

Fortunately the answers are the same :)

Melody Jul 19, 2017