Can someone help me with vectors?

The position vector of a fleet footed African cheetah is given by the following funcion of time: x(t)=4.0m-(2m/s)t-(3m/s^2)t^2





a) what is the position vector of the cheetah at the instant the stopwatch commences, t=0s?


b) what is the position vector of the cheetah when the stopwatch indicates t=3s?


c)what is the change in position vector during the three seconds interval between these time instants?

Can someone help me with vectors?
Not really much of a vector problem.


Just insert the value of t and solve.





The first term (4.0) is the initial position of the cheetah


The second term is the velocity of the cheetah as a function of time


The third term is the acceleration of the cheetah as a function of time





With t = 0.0, the position of the cheetah is simply 4.0 meters.


When t = 3 you have


Position = 4 - 2*3 - 3*3^2 = 4 - 6 - 27 = -29


And the change in position is simply 4 - (-29) = 33 meters
Reply:you sure those minus signs aren't really + signs?
Reply:Stuff t=0 into the equation and you get 4.0M. Do the same with 3 to get that answer. Take the difference to get the third. Look out for the fact that the signs indicate the cheetah is running in the negative x direction.