d^2x/dt^2 = 2 t - 1
dx/dt = t^2 - t + c
at t = 0, dx/dt = 2 so c = 2
so
v = dx/dt = t^2 - t + 2
x = (1/3) t^3 - (1/2) t^2 + 2 t + c
when t = 0 x = a so c = 1
x = (1/3) t^3 - (1/2) t^2 + 2 t + 1
so at t = 6
v = 36 -6 + 2 = 32 m/s
and
x = 67
total distance = 67-1 = 66
assuming x is always positive (sure looks that way
The acceleration of a perticle as it moves along a straight line is geven by a=(2t-1) m/s2. If x=1m and v=2m/s when t=0s, determine the particle's velocity and position when t=6s. Also, determine the total distance the particle travels during this time period.
2 answers
Thanks