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Uniformly Accelerated Motion

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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
+26814
+3

Use v^2 = u^2 + 2as

v = 0 m/s. Final velocity

u = 80000/3600 m/s. Initial velocity

s is distance travelled in metres.

s = (80000/3600)^2/(2*7.5) m → 32.9 m

.

Alan  Jul 19, 2017
#1
+26814
+3

Use v^2 = u^2 + 2as

v = 0 m/s. Final velocity

u = 80000/3600 m/s. Initial velocity

s is distance travelled in metres.

s = (80000/3600)^2/(2*7.5) m → 32.9 m

.

Alan  Jul 19, 2017
#3
+1

Guest Jul 20, 2017
edited by Guest  Jul 20, 2017
#4
+1

And why did + in your formula become / in your solution? Thanks. :]

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
+92888
+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
edited by Melody  Jul 20, 2017