the distance s(t) = 12t - t^2 + c
for some c.
Since s(1) = -10, c=-21, and so
s(t) = -21 + 12t - t^2
a = dv/dt = -6t
so just plug in t=4
plug in t= 0 and 10 to find where the particle is at each time, then get the displacement by evaluating s(10)-s(0)
A particle travels along a straight line with a velocity v=(12-3t^2)m/s.where t is in the seconds. When t=1s.the particle is located 10m to the left of the origin. Determine the acceleration when t=4s.the displacement from t=0s to t=10s.and the distance the particle travels during this time period.
1 answer
1 answer