Question

If v(t) = 2t + 1, the velocity of an object over time, what is the acceleration of the object at t = 1?

Answer 1

@rajathsbhat can u help?

Answer 2

do you know how to differentiate functions?

Answer 3

no

Answer 4

i don't see how else you could answer this question...
i'll teach you some basics you need to solve this problem:
1: \(\Large \frac{d(t^{n})}{dt}=n*t^{n-1}\)
2: \(\Large \frac{d(k)}{dt}=0\)
3: \(\Large \frac{d(k * t^{n})}{dt}=k*n*t^{n-1}\)

You should use these rules to differentiate the velocity because a=dv/dt.
Tell me what you get.

Answer 5

okay hold on

Answer 6

so t = 1 what is d, k? and n?

Answer 7

k is a constant, like 1 or 2 or whatever.

Answer 8

ok

Answer 9

n is the power to which t is raised. In this case, it's would be 1.
and you should substitute t=1 only AFTER you've differentiated the function.

Answer 10

*it

Answer 11

ok

Answer 12

what is d?

Answer 13

it's called a differential. it's just a notation.
it's looks like this: \[\large \frac{d}{dt} (whatever)\]

Answer 14

ohh ok

Answer 15

so will it be d(1^1)/d(1) = 1 x 1^-1

Answer 16

no you only substitute for t after you're done with the diffentiation.
so it's like \[\large \begin{align}\frac{d}{dt}(2t+1)&=\frac{d}{dt}(2t)+\frac{d}{dt}(1)\\\\
&=2\times1\times t^{1-1}+0\\\\
&=2.\end{align}\]

Answer 17

not that you don't have t anymore. that means that the acceleration doesn't depend on the time. I.e. you have constant acceleration.

Answer 18

IF you ended up with something like 2(t) in the end, you'd then substitute t=1 or 2 or anything that your question asks to get the acceleration at that time.