How Long to Accelerate for Desired Time?

[pointintime]

Ok well I got down to this and don't know what to do please someone help me...

-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

 

[rl.bhat]

Ok well I got down to this and don't know what to do please someone help me...

-2^-1 t3^2 - 180 s t3 + t3^2 = a^-1 (Vo (180 s) - 1,100 m + Xo)

You can solver this problem very easily by drawing time- velocity graph.
You can find the required time by considering the area above the last time interval i.e. 180 seconds.
This area consists of a rectangle with area 5.494*180 = 988.92 m.
The remaining area = 1100 - 988.92 = 111.1 m.
This distance is equal to the area of the trapezium with parallel sides 180 s and t2 s, and height 0.2*t1 .
So 111.1 = 1/2*0.2*t1(180 + t2) Since t1 + t2 = 180 s, t2 = (180 - t1)
Hence 111.1 = 1/2*0.2*t1(180 + 180 - t1)
Or
111.1 = 1/2*0.2*t1(360 - t1)
Solve this quadratic to find t1.

 

[pointintime]

um 5.494 is the intial velocity of the person running before the acceleration occurs in which an acceleration is present and at the end of 180 second the person is no longer running in fact he stops accelerating before the 180 seconds, the end of the race, the question asked for how much time he msut accelerate in order to get under 30 min

 

[pointintime]

So I don't see how your way works

 

[rl.bhat]

So I don't see how your way works

During 180 seconds he accelerates for t1 seconds and runs uniformly for t2 seconds. Both these are taken into account in the area of the trapezium.

You solve the quadratic for t1 and check the answer.

 

[kuruman]

There is absolutely no reason to have x₀ in your expression.

Step by step help

1. Go back to posting#45 by Jotun.uu.
2. Read it very carefully.
3. In the equation shown in #45, enter the numbers your have for the acceleration a = 0.2 m/s². v = 5.494 m/s and s = 1100 m
4. Solve the quadratic equation for t. If you are not sure how to do this, google "quadratic formula".

 

[pointintime]

Ok I graphed into a graphic calculator

y1 = -2 X^2 - 180 x + X^2

y2 = .2^-1 (5.494*180-1100)

and got this

(3.111, -555.4)

I assume I need the 3.111 seconds for my answer
but sense this is physics not algebra how do I rearange for y3?

 

[pointintime]

wait so I just forget

.2^-1 (5.494*180-1100)

that part of the equation and just do the opposite of b square root... yada yada yada?

 

[kuruman]

The correct answer is 3.11 s. The negative answer is unphysical. I don't know what y3 means.

 

[pointintime]

x = (-b +/-sqrt(b^2 - 4ac))/2a ("sqrt" = square root)

so is the 2a part...

the a is -2^-1 so does it become part of numerator and I get this?
x = a(-b +/-sqrt(b^2 - 4ac))/2 ("sqrt" = square root)

 

[pointintime]

I don't know how to get this mess into standard form...

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)

 

[pointintime]

I have to get it into

y = ax^2 + bx + c

don't know how

 

[pointintime]

rewrote for zero don't know what to do

-2^-1 t3^2 - 180 s (t3) + t3^2 = a^-1 (Vo (180 s) - 1100 m + Xo)
(Vo (180 s) - 1100 m + Xo)^-1 a(-2^-1 t3^2 - 180 s (t3) + t3^2) = 0

:O