Question

suppose that the position of a particle as a function of time is given as:
x(t)=1t^4(i^hat)+(-2t^3+1t^4)j^hat
a) determine velocity as a function of time:
b)determine acceleration as a function of time:
c)determine the magnitude of the velocity at t=0.7s

Answer 1

x = t^4 i + (-2t^3+1t^4) j ????
if so
dx/dx = 4 t^3 i + (-6 t^2 + 4 t^3) j = velocity
d^2x/dt^2 = 12 t^2 i +(-12 t + 12t^2) j = 12 [ t^2 i + (t^2-t) j = acceleration
at t = 0.7
dx/dt = (4 * 0.7^3) i + (4 * 0.7^3 - 6 *0.7^2) j
= 1.372 i + (1.372 - 2.94) j
= 1.372 i - 1.57 j
|v| = sqrt (1.37^2 + 1.57^2)