Posted by Hannah on Saturday, February 26, 2011 at 10:25pm.
This is an initial value problem, where you would solve a differential equation (or an integral), and with a given set of initial values, determine the integration constant.
Let x(t)=position at time t
v(t)=tsin(t^2)
∫v(t)dt = ∫tsin(t²)dt
Integral of velocity gives distance
For the right-hand side, substitute u=t^2,
du=2tdt
x(t)=∫(1/2)sin(u)du
=-cos(u)/2+C'
=-cos(t^2)+C
Given x(0)=3
=>
x(0)=-cos(0)+C = 3
C=3+cos(0)=4
=>
x(t)=-cos(t^2)+4
x(2)=4-cos(2^2)=4-cos(4)
Distance travaelled = x(2)-x(0)=?
Hannah, Anastasia, Carmen,Yuka, Lena, Catalina, Jamima, ...
I appreciate your inventiveness, but please, stop inventing screen names.
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