Learn how to calculate velocity with S>V>T help and practice questions

[spam_hammer]

Hi all,
This is my first post so I hope this doesn't break any rules, to the best of my knowledge it doesnt, but let me know.

i am curently revising for a maths exam and i am practising questions on velocity.

how should i go about calculating a questoin in the form

v= e^t

at what value of t does the particle have a velocity of x ?


I have removed the numbers so that you guys arent answering the actual question for me. I am unsure if the question can still be answered as i have written it , if not i will post the actual figures.


also just to check an answer I have calculated ( I think):

find the acceleration of a particle whose velocity v=e^3t-2 at t=1

My answer acceleration T= 1 m/s^2

thanks

 

[JG89]

For the first question, if v denotes velocity, then you have v = e^t. You want to know when v, or e^t, is equal to x. So you have x = e^t. Use the natural logarithm to solve for t.

For your second question, your answer is wrong. If v = e^(3t) - 2, then you know that the acceleration = dv/dt. Calculate dv/dt at t = 1.

 

[spam_hammer]

okay so dv/dt at t=1

v=e^3t-2 t=1 so v= e¹

differentiation = 1e⁰= 1

I'm afraid I am a maths dunce so i can't see where i went wrong

For the first question, if v denotes velocity, then you have v = e^t. You want to know when v, or e^t, is equal to x. So you have x = e^t. Use the natural logarithm to solve for t.

I am unsure how to use ln here?

apologies for my simpleness of mind :P

 

[JG89]

v = e^(3t - 2)

Differentiate using the chain rule. Then just plug in 1 wherever you see t.

For the second question if x = e^t, then ln(x) = ln(e^t) = tln(e)?

What's ln(e) equal to?

 

[spam_hammer]

i am going to say, tentatively, 1 ?

 

[ImAnEngineer]

v = e^(3t-2)
First take the derivative, then plug in t=1.
dv/dt = 3e^(3t-2) (chain rule)
t=1
dv/dt = 3e

v=e^t
ln v = lne^t = tlne = t
So: t = lnv

 

[spam_hammer]

Evening all,

another day, another revision question in the endless battle against my brain to make it understand chain rule.

this time I think I have worked through it and would much apreciate it if someone would notify me of the myrriad mistakes with which it is ridden i am sure.

okay question :
The velocity v in m s^-1 of a moving body at time tseconds is given by v=e^2t-1 when t=0.5 displacement equals 10m. Find the displacement when t=1


My workings :

Chain rule v= e^u where u= 2t-1


intergrate 2t-1 = t²-1t+c


when t = 0.5 displacement (a) = 10

Then using this to find out value for c:

((0.5)²-0.5+c) *e ^2(0.5)-1=10

(-0.25+c)*e⁰=10

(-0.25+c)* 1 = 10

-0.25= 10 -c

-10.25= c

substituiting this information back into the question



a=((1)² - 1(1) -10.25)* e ^2(1)-1

-0.25 * e = -27.9



the first thing that makes me think i am wrong is the negative value for displacement .



Im not sure if any of this even makes sense. please let me know

 

[ImAnEngineer]

Evening all,

another day, another revision question in the endless battle against my brain to make it understand chain rule.

this time I think I have worked through it and would much apreciate it if someone would notify me of the myrriad mistakes with which it is ridden i am sure.

okay question :
The velocity v in m s^-1 of a moving body at time tseconds is given by v=e^2t-1 when t=0.5 displacement equals 10m. Find the displacement when t=1My workings :

Chain rule v= e^u where u= 2t-1intergrate 2t-1 = t²-1t+c

Looks like you have to calculate the definite integral!

The anti-derivative of e^2t-1 is (1/2)e^2t-1.
You probably don't understand why, so that's why I'll explain. If you take the derivative of, say: e^ax, the derivative is ae^ax (chain rule- a is the derivative of ax). When you take the anti-derivative, you have to 'undo' this extra factor a that comes up front, so you divide by it. Suppose we have again e^ax, the anti-derivative will be (1/a)e^ax, because if we differentiate it again we get back to e^ax.

So what you have to calculate is:

int₀¹[(1/2)e^2t-1] dt

Remember that you don't have to find the constant when taking a definite integral (this makes physical sense as well; it only matters how much distance has been traveled in the time interval from 0 to 1, not how much had been traveled before or after, which would basically be the meaning of having a constant in this case. Hope this doesn't confuse you. Forget about it if you don't get it )