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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

Best Answer 

 #1
avatar+25995 
+3

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
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6+0 Answers

 #1
avatar+25995 
+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
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+1

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

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

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

Guest Jul 20, 2017
 #5
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+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
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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.     

-------

 

 

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
avatar+89804 
+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.    laugh

 

 

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

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