- #1
positive infinity
- 9
- 0
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??!
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??!