What is Chelsea's kinetic energy at t=2.50s?

[positive infinity]

Hello again and here is another problem I can't seem to figure out.

Chelsea, a calico cat, runs in a straight line chasing a car down the street. Chelsea weighs 36.5 N. and her position along the road is given by
x(t)=(3.50m/s)t+(0.425m/s^2)t^2+1.00ms/(t+1.00s)
Calculate Chelsea's kinitec energy at t=2.50s

The answe is 57.1 joules

Okay not I am not to sure if I'm supposed to find the first or the second derivative. well I compiled it like so
x(t)=at+bt^2+c/(t+d)
x'(t)=a+2bt+(t+d)*F((x)c)-c*F((x)t+d)/(t+d)^2 which comes out to
x'(t)=3.50m/s+2(0.425m/s^2)(2.50s)+(2.5m/s^2)/(2.5m/s+1.00s)^2 = 5.83m
(m*N)=Joules, 5.83m*36.5 N = 212.8 J <--way off so i tried for a second deriv.
x''(t)=2(0.425m/s^2+6.25m/s^2/(2.5s*1.00s)^3= 0.996m 0.996m*36.5N=36.3 J <-- still wrong , Where am I messing up can anyone please help me??!

 

[Curious3141]

What is the formula for kinetic energy in terms of mass and velocity ?

Is velocity the first differential or second differential of the position wrt time ?

The differentiation is wrong. What is $$\frac{d}{dx}(\frac{1}{x + c})$$ ?

 

[positive infinity]

well I used the quotient rule, at the end because it seemed most logical. all we have to do is calculate the amount of joules whic m*N= J and we find the deriv. of the position time function given But, I'm not so sure.

 

[Curious3141]

well I used the quotient rule, at the end because it seemed most logical. all we have to do is calculate the amount of joules whic m*N= J and we find the deriv. of the position time function given But, I'm not so sure.

From the way you wrote the expression out, isn't it of the form :

$$3.5t + 0.425t^2 + \frac{1}{t + 1}$$

rather than

$$\frac{3.5t + 0.425t^2 +1}{t + 1}$$ ?

The former expression does not need quotient rule to differentiate it. Well you can, but there's a much easier way to do it. But first clarify the question. Besides where do all the F(x) etc come from ?

 

[tanky322]

The former expression does not need quotient rule to differentiate it. Well you can, but there's a much easier way to do it. But first clarify the question. Besides where do all the F(x) etc come from ?

$$3.5t + 0.425t^2 + \frac{1}{t + 1}$$ That is the correct equation.

Im in the same class, by the F(x)'s that's the position time function we are givin. Now I am confused on how to find Kinetic energy. Do I just determine the first derivative? Will that give me K.E(t)?

Thanks,
Andy

 

[Curious3141]

$$3.5t + 0.425t^2 + \frac{1}{t + 1}$$ That is the correct equation.

Im in the same class, by the F(x)'s that's the position time function we are givin. Now I am confused on how to find Kinetic energy. Do I just determine the first derivative? Will that give me K.E(t)?

Thanks,
Andy

The first derivative of the displacement will give the velocity.

The kinetic energy is given by 0.5*mass*(velocity)^2

The mass can be found from the weight of the cat (given).

Can you differentiate $$\frac{1}{t + 1}$$ wrt t

by chain rule ? Or you can use quotient rule, but chain rule is easier here.