Question

Is there anyone that can help me with calculus?

Answer 1

I need help finding velocity

Answer 2

The displacement of a weighted spring from its equilibrium on the time interval [0, 30] is modeled by the position function y(t) = 5 cos 6t, where y is measured in meters and t in seconds. Find the velocity of the weight the first time its acceleration is zero.

I know that velocity is finding the first derivative. For this one do I find the derivative and then set it equal to zero?

Answer 3

Acceleration is the second derivative, y''.
Find y'', set it to 0, find t.
Put that t in y' and calculate y'

Answer 4

So I have to find the second derivative then set it to zero. Okay one minute let me do that.

Answer 5

-30sin(6t) is the derivative right?

Answer 6

That is the first derivative (or velocity)
Find the derivative of velocity to get acceleration.

Answer 7

-180cos(6t) is the second derivative.

Answer 8

set it to 0 and solve for t.

Answer 9

3pi/2?

Answer 10

I really hate trig functions. It makes things harder. Haha.

Answer 11

The cosine function starts at 1 and first becomes zero at pi/2
So cos(6t) will become zero when 6t = pi/2 or t = pi/12.

Answer 12

pi/12

Answer 13

Now I plug pi/12 into the first derivative?

Answer 14

put t = pi/12 in the velocity equation (that is the first derivative) and calculate the velocity.

Answer 15

-30

Answer 16

Yes, -30 meters / second.

Answer 17

-30 m/s

Answer 18

That was easier than I thought. Sometimes it is hard when they are in word problems to figure out what to do.

Answer 19

Could you help me with one more?

Answer 20

Post it here and I will revisit as I am answering another question. brb.

Answer 21

A person walks from a point of origin with the position function x(t) = 2 - 2|t - 1|. Find the total distance traveled by the person on the interval 0 ≤ t ≤ 5.

Answer 22

So do I find the derivative first?

Answer 23

No need for derivatives here.
The function x(t) tells you the location of the person at time t.
We need to get rid of the absolute bars.
So write the function as two parts: 1) when what is inside the absolute bar is negative and 2) when it is positive.

Answer 24

when t=1 it is zero
when t=-1 it is negative
Is that what I am supposed to do?

Answer 25

My worksheet has these to choose from.
6
8
10
11

Answer 26

We don't go negative time.

Yes, when t = 1, the quantity inside the absolute bar goes to zero.
Let us create a table:
t = 0, x = 0
t = 1. x = 2
t = 2, x = ?
complete the table till t = 5

Answer 27

we are solving for inside the bar right?

Answer 28

or the whole equation?

Answer 29

Here since the interval is small with t = 0, 1, 2, 3, 4 and 5 it would be easier to just go with the whole equation.
x(t) = 2 - 2|t - 1|
put t = 0, then x = 0
t = 1, x = 2
t = 2, x =
t = 3, x =
t = 4, x =
t = 5, x =

Answer 30

t = 2, x = 0

Answer 31

t=2 x=0

Answer 32

Do I just put in the numbers for t?

Answer 33

Yes. Just put t = the number and find x(t).

Answer 34

2-2|3-1|=2

Answer 35

when t = 3, x = 2 - 2|3 - 1| = 2 - 2(2) = 2 - 4 = -2

Answer 36

t=4 x= 2-2(3)=2-6=4

Answer 37

-4

Answer 38

t=5 x=2-2(4)=2-8= -6

Answer 39

yes.

Answer 40

Distance traveled would be 8 right?

Answer 41

wait there is an interval

Answer 42

t x
0 0
1 2
2 0
3 -2
4 -4
5 -6

The person starts at x = 0, walks in the positive x direction to x = +2, turns around and walks back to the starting position x = 0 and then walks further in the negative x direction to another 6 units. So the person has walked: 2 + 2 + 6 = 10.

Answer 43

oh I see