using calculus to solve problem: distance x= 9/2t + t meters at time t >= 1

[bumblebee123]

Hi,
So I have this problem:
A particle moves along a line.
For t (more than/equal to)1, the distance of the particle from O at time t is x meters, where:

x= 9/2t + t

Find the acceleration of the particle when t = 3

I have been taught to differentiate to find velocity, so dx/dt an then use that to find acceleration dv/dt

I was having trouble finding inverse of 9/2t so that I could differentiate, I came up with :

18t^-1 however when I work with it as suggested above my answer for acceleration was 1 1/3

I also managed to get 2/3 m/s^2 can't find what I used. In anycase neither answers are correct.

The correct answer is 1/3 m/s^2

Please can some one show me how to get to the answer. I wonder if my problem lays in finding the inverse in the first place.

Thank you in advance for your help.

 

[HallsofIvy]

Hi,
So I have this problem:
A particle moves along a line.
For t (more than/equal to)1, the distance of the particle from O at time t is x meters, where:

x= 9/2t + t

Please use parentheses! Is this (9/2)t+ t or 9/(2t)+ t or 9/(2t+ t)? Since the first would be more simply written (11/2)t and the third 3/t, it is most likely the second 9/(2t)+ t.

Find the acceleration of the particle when t = 3

I have been taught to differentiate to find velocity, so dx/dt an then use that to find acceleration dv/dt

I was having trouble finding inverse of 9/2t so that I could differentiate, I came up with :

18t^-1 however when I work with it as suggested above my answer for acceleration was 1 1/3

Assuming this was indeed 9/(2t) then you can write it as (9/2)t^-1. The derivative of x(t)= (9/2)t^-1+ t is
v(t)= (-9/2)t^-2+ 1 and the acceleration is a(t)= 9t^-3. At t= 3 that is 9/27= 1/3.

I also managed to get 2/3 m/s^2 can't find what I used. In anycase neither answers are correct.

The correct answer is 1/3 m/s^2

Please can some one show me how to get to the answer. I wonder if my problem lays in finding the inverse in the first place.

Thank you in advance for your help.

You simply use 1/t= t^-1. The "9/2" is simply a number and you don't have to do anything to it.

 

[bumblebee123]

Sorry about that. It should read 9/(2t) + t

Actually I probably need to show it like this:

[9/(2t)] + t

so its 9 divided by 2t and then you add t

Sorry I have trouble writing it. I hope that comes across right

Thank you Hallsofivy, just read through - I understand now.