x(t) = 2.17+4.80t^2−0.100t^6
v(t) = 2.17+ 9.60t - 0.600t^5
a(t) = 9.6 - 3.0t^4
As you say, v(2) = 0, so a(2) = -38.4
The position of the front bumper of a test car under microprocessor control is given by x(t)=2.17m+(4.80m/s2)t2−(0.100m/s6)t6.
Find it's acceleration the second instant when the car has zero velocity.
I already know that the position of the car when it has zero velocity for the second time in x=15 m but that's as far as I could get.
3 answers
how did you find that x = 15?
x=15
t=2s
x(t)=2.17+4.80t^2-0.100t^6
x(2)=2.17+4.80(2)^2-0.100(2)^6
x(2)=14.97m or 15m
t=2s
x(t)=2.17+4.80t^2-0.100t^6
x(2)=2.17+4.80(2)^2-0.100(2)^6
x(2)=14.97m or 15m
1 answer