calculus type question (don't know what topic it is from though) (2 Viewers)

[click here]

easy motion question

If a = 2t2 + 1 and v = 2 and x = -1 when t = 0, find x when t = 4.

Enter your answer as a decimal correct to the nearest integer.

[ANSWER: 58]

2. If v = x + 2 and x = 0 when t = 0, find x when t = 3.

[ANSWER: 38.17]

Thanks

 

[hyparzero]

Re: easy motion question

Looks like differential equations

1. a = 2t^2 + 1 and v = 2 and x = -1 when t = 0, find x when t = 4.

a = 2t^2

=> dv/dt = 2t^2 ..................multiply by dt each side
=> dv = 2t^2 dt ....................integrate

=> v = t^3 + c ...................v = 2 when t = 0 hence c = 2

=> v = t^3 + 2
=> dx/dt = t^3 + 2

=> dx = t^3 + 2 dt

=> x = 1/4 t^4 + 2t + C ................ when t = 0 x = -1

Hence C = -1

Therefore x = 1/4 t^4 + 2t -1 ............then substitute t = 4 to find x.



2. Similar to 1. Just remember v= dx/dt = x +2

dx/(x + 2) = dt ............ integrate both sides

ln(x + 2) = t + C ................. x = 0 when t = 0; therefore C = ln2
=> ln(x + 2) - ln2 = t

therefore

t = ln [ (x + 2) / 2 ]..................substitute t = 3 to find x (use exponentionals)

 

[click here]

Re: easy motion question

Thanks a lot!

But I think you got the first question wrong

 

[hyparzero]

Re: easy motion question

Thanks a lot!

But I think you got the first question wrong

yup, didnt read ur question properly, i missed the +1

 

[click here]

Re: easy motion question

One last question:

If a particle, initially at rest, has velocity given by v = t(2-t).

Find the distance travelled by the particle between t = 0 and t = 8.

Enter your answer as a decimal correct to 2 decimal places in the box below.



THANKS!

 

[hyparzero]

Re: easy motion question

v = t(2-t)
=> v = 2t - t²
=> dx/dt = 2t - t²

Therefore dx = (2t - t²)dt

Hence ∫dx = ∫(2t - t²)dt

=> x + k = t² - t³/3 ......where k is a constant

When t = 0, x = 0, hence k = 0

Therefore x = t² - t³/3

Let t = 8
Hence x = 8² - 8³/3

=> x = 64 - 170.6666666667

Hence x = -106.6666666667 metres

 

[click here]

Once again thanks for your help

But the answer says its 109.33

 

[Riviet]

We are finding distance after 8 seconds, not displacement. Notice that particle changes direction and has velocity 0 at t=2, so when integrating, we must split the integral into two parts:
2XXXXXXXX8
∫ 2t-t²dt +|∫ 2t-t²dt|
0XXXXXXXX2

The absolute value is taken since distance can't be negative [but displacement can be negative]. Hope that helps.

 

[click here]

Thanks to the always helpful Riviet!

But I tried doing it your way and I got the answer of -58.67

 

[SoulSearcher]

After all that integration and substitution, you should get:
(4-8/3) + |[(64-512/3) - (4-8/3)]|
= 4/3 + |(-324/3)|
= 4/3 + 108
= 109.33, as required.

 

[hyparzero]

meh, i suck