# S>V>T help with velocity

Hi all,
This is my first post so I hope this doesnt 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

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.

okay so dv/dt at t=1

v=e3t-2 t=1 so v= e1

differentiation = 1e0= 1

I'm afraid I am a maths dunce so i cant 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

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?

i am going to say, tentatively, 1 ?

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

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=e2t-1 when t=0.5 displacement equals 10m. Find the displacement when t=1

My workings :

Chain rule v= eu 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

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=e2t-1 when t=0.5 displacement equals 10m. Find the displacement when t=1

My workings :

Chain rule v= eu where u= 2t-1

intergrate 2t-1 = t2-1t+c
Looks like you have to calculate the definite integral!

The anti-derivative of e2t-1 is (1/2)e2t-1.
You probably don't understand why, so that's why I'll explain. If you take the derivative of, say: eax, the derivative is aeax (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 eax, the anti-derivative will be (1/a)eax, because if we differentiate it again we get back to eax.

So what you have to calculate is:

int01[(1/2)e2t-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 travelled 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 )

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