Question

The position x, in meters, of an object moving along the x-axis is given as a function of time t, in seconds, as x(t)=0.0981t^4-0.231t^3+0.773t^2+5.83t-9.27. Find the object's acceleration at time t=3.45s

Answer 1

\(\huge \sf Acceleration = \huge \frac{d(velocity)}{dt}\)

Answer 2

So first we will have to calculate the velocity as we calculated in the last question. Can you calculate the equation for velocity ?

Answer 3

\(\sf \huge Velocity = \huge \frac{dx}{dt}\)

Answer 4

would the 9.27 be negative? when subtracting it from .585?

Answer 5

\(\sf = 0.3924t^3-0.693t^2+1.546t\)
This is velocity

Answer 6

Now u'll have to differentiate it once more to get an equation for acceleration. Can you ?

Answer 7

ok so,-0.231t^3+0.773t^2+583t = the velocity?

Answer 8

Wr did you got this from ?

Answer 9

from the problem

Answer 10

or is x the total velocity?

Answer 11

x is position/Displacement

Answer 12

You differentiate it to get velocity.

Answer 13

Then differentiate velocity to get acceleration.

Answer 14

You know how to differentiate ?

Answer 15

not exactly

Answer 16

Haven't u learnt about calculus ?

Answer 17

i've just started calculus, we are currently in limits

Answer 18

ok, i am simplifying it for you though u'll learn it in detail

Answer 19

ok,thanks

Answer 20

If i say you to differentiate \(\sf 5t^n\) with respect to t, then the answer will be \(\sf 5nt^{n-1}\)

Answer 21

Example:
Differentiate \(\sf 5t^7\) with respect to t
Ans: \(\sf 35t^6\)

Answer 22

Got it ?

Answer 23

so you multiplied the 7 to the 5 and subtracted 1 from the 7?

Answer 24

yes

Answer 25

Now differentiate \(\sf 8t^9\) wrt t

Answer 26

72t^8

Answer 27

\(\huge\checkmark\)

Answer 28

Differentiation of a constant term is always zero

Answer 29

Example if i say you to differentiate 36 wrt t, then the answer will be 0.
Got it ?

Answer 30

ok, yes...if the number is negative, for example -4t^3, would the 4 remain negative or should it be positive?

Answer 31

It will remain negative

Answer 32

Now,
Differentiate \(\sf 3t^7-5t^4+68\) wrt t

Answer 33

21t^6-20t^3+68

Answer 34

68 will become 0 after differentiation. It's a constant term

Answer 35

oh ok, i see

Answer 36

One more time,
Differentiate \(\sf 45x^4+78x^3-8.9t+56\) wrt x

Answer 37

180t^3+234t^2-8.9t

Answer 38

\(\sf 8.9x\) can be written as \(\sf 8.9x^1\). So when you will differentiate it wrt to x it will become 8.9

Answer 39

ok, i was unsure about that

Answer 40

also you can not change x with t

Answer 41

oh sorry i didn't realize i did that...

Answer 42

Answer will be \(\sf 180x^3+234x^2-8.9~and~not~180t^3+234t^2-8.9t\)

Answer 43

are you sure you can do the next one correctly in one attempt ?

Answer 44

yeah

Answer 45

allryt let's see
Differentiate \(\sf 0.0981t^4-0.231t^3+0.773t^2+5.83t-9.27\)

Answer 46

0.3924t^3-0.693t^2+1.554t^1+5.83t

Answer 47

\(\sf 0.3924t^3-0.693t^2+1.546t^1+5.83\)

Answer 48

how does the t on 5.83 cancel out?

Answer 49

Differentiation of 5.83t = ?

Answer 50

5.83

Answer 51

yes, 5.83t = \(\sf 5.83t^1\), when u differentiate it, \(\sf 1\times 5.83t^{1-1}=5.83t^0=5.838\)

Answer 52

so from .3924t^{3}-.693t ^{2}+1.544t+5.83 you plug in the time 3.45s for t?
.....oh yes ok i see, i'm sorry my mistake

Answer 53

No, you need to differentiate it once more

Answer 54

Differentiate
\(\sf 0.3924t^3−0.693t^2+1.546t^1+5.83\)

Answer 55

1.1772t^2-1.386t+1.544

Answer 56

how do you know when you are done differentiating? is it when it gets into slop-point form?

Answer 57

or is it when t is simplified?

Answer 58

When you differentiate displacement you get velocity. When you differentiate velocity you get acceleration.
So if you have to calculate acceleration from displacement you need to differentiate it twice. Got it ?

Answer 59

First differentiation will give you velocity and the second one will give acceleration. Ryt ?

Answer 60

oh ok...

Answer 61

so it should be 1.1772t^2-1.386t+1.544 and plug in 3.45 for t right?

Answer 62

yes

Answer 63

thank you so much!!!