At time t = 0 s, an object is observed at x = 0 m; and its position along the x-axis follows this expression: x = –4t + t2, where the units for distance and time are meters and seconds, respectively. what is the object's displacement between t = 1.0 s and t = 3.0 s?

The object displacement is 0 meters.

Explanation:

Because the expression for the object position as function of time is:

At t= 3.0=t2 seconds the x-position of the object is:

and at t=1.0=t1 seconds the x-position of the objects is:

The total displacement () of the object between t2 and t1 is the difference between x(3.0s) and x(1.0s):

The displacement of the object is zero!!!

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

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