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where I'm doing wrong? (math)


Vastor

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Hey guys,

 

simple_algebra_problem.png

based on the picture above, the answer given by my teacher :-

 

[math] 22 = \frac{285 + \frac{t-12}{2} * 30}{t} [/math]

 

[math] 22t = 285 + 15t - 180 [/math]

 

[math] 7t = 105 [/math]

 

[math] t = 15 [/math]

 

 

but for my answer:-

 

[math] 22 = \frac{180 + \frac{t-9}{2} * 40}{t} [/math]

 

[math] 22 = 180 + 20t - 180 [/math]

 

[math] 2t = 0 [/math] ??!

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I guess along these lines: an object accelerates from 0m/s at time 0s to 40m/s at time 9 sec at constant acceleration, it then decelerates (again constant) back to 0m/s at time t sec - at time 12s the velocity is 30m/s so the velocity at time=12sec is 30 m/s, it then decelerates such that at time = t sec the velocity = 0m/s

 

 

If the average velocity is 22m/s what is t?

 

Distances travelled (average speed each section multiplied by time of each section)

 

[math] \frac{0+40}{2}*(9-0) + \frac{40+30}{2}*(12-9) + \frac{30+0}{2}*(t-12) [/math]

 

Average Speed = Distance travelled per above Divided by Total Time.

 

Find t.

 

Ewmon is completely correct you need to be more careful in your algebra. And you need to get your diagrams right

 

edit - brain fart

Edited by imatfaal
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The question is, "Calculate the value of t, if the average speed for the t seconds is 22ms"

 

for imatfaal, I don't know how this calculation become wrong...

 

simple_algebra_problem2.png

If that is a triangle (in other words: if your acceleration is constant) then the 2nd blue line crosses the t-axis at t=21 s. It's a simple extrapolation of the slope:

 

For the decelleration, using your picture as input:

[math]v(t) = 40 - \frac{(40-30)}{(12-9)}\cdot{(t-9)}=40 - \frac{10}{3}\cdot{(t-9)}[/math]

So, that gives [math]v = 0[/math] at [math]t = 21[/math]

 

However, if your question is: "Calculate the value of t, if the average speed for the t seconds is 22m/s", we have a problem.

 

Because if the acceleration is constant, and the top speed is 40 m/s, then the average speed should logically be 20 m/s... not 22 m/s.

So, that would imply that your picture is wrong (it's not a triangle, but some other kind of graph, and the lines should probably not be straight). Also, it means we do not have enough information to solve this problem.

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If that is a triangle (in other words: if your acceleration is constant) then the 2nd blue line crosses the t-axis at t=21 s. It's a simple extrapolation of the slope:

 

For the decelleration, using your picture as input:

[math]v(t) = 40 - \frac{(40-30)}{(12-9)}\cdot{(t-9)}=40 - \frac{10}{3}\cdot{(t-9)}[/math]

So, that gives [math]v = 0[/math] at [math]t = 21[/math]

 

However, if your question is: "Calculate the value of t, if the average speed for the t seconds is 22m/s", we have a problem.

 

Because if the acceleration is constant, and the top speed is 40 m/s, then the average speed should logically be 20 m/s... not 22 m/s.

So, that would imply that your picture is wrong (it's not a triangle, but some other kind of graph, and the lines should probably not be straight). Also, it means we do not have enough information to solve this problem.

 

heh, that's answered all of the question. and the picture is the "question picture" itself that I draw on paint. (yes, I sure the picture show straight line => constant ac/deceleration)

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vastor: a picture isn't very useful without the question itself.

are you looking for t?

then the slope of the line is all you need.

(40 -30)/(9-12) = -10/3

 

40 -10 = 30

9 +3 =12

30 -10 = 20

12 +3 =15

20 -10 = 10

15+3 = 18

10 -10 = 0

18+3 = 21

21 = t

or more simply 40/10 = 4, 4*3 +9 = 21.

 

i'm not quite sure what you're trying to do with the area of a triangle aproach. i see your logic in the equation, and you're correct; area of left triangle + area of right triangle = area of whole triangle. i would need to see your teacher's logic as well though to understand what's going on.

 

 

 

 

 

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