Recall

We were discussing kinematics:  knowing the position as a function of time, x(t), the velocity and acceleration are obtained by taking the appropriate derivatives:
 

v(t) = dx/dt = [xdot]   and   a(t) = dv/dt = [vdot] = d²x/dt² = [xddot].
 

Constant Acceleration

Suppose we know that the acceleration of an object is constant, a = constant = a0 .  This is the case for an object falling under the influence of gravity, neglecting friction.  Can we determine the velocity, v?

Since  a = dv/dt  then  v(t) = [Int from t0 to t] a0dt  +  constant.  Why is there an additional constant?  This is easily seen by letting t = t0, in which case the integral becomes zero.  But it is not required that the velocity at t = t0 is zero.  (In fact, we can let t0 be any time at all, so if the velocity were zero at t0, it would have to be zero at all times!)  It is also readily apparent that the constant is the velocity at time t0.  For the remainder, let us assume t0 = 0, and the velocity at t=0 is v0.  Then, the velocity is v(t) = [Int from 0 to t]a0dt + v0 = a0t + v0 .  This result should look familiar.

Now suppose that we want to know the position as a function of time.  Using the same technique as above, we can integrate the velocity (now known) to determine the position:  x(t) = [Int from 0 to t] v(t)dt + constant.  By the same argument used above, the constant is the postion at time t=0, which we will call  x0 .  This time, the integral is slightly more difficult, but still quite straightforward:  x(t) = [Int from 0 to t](a0t + v0)dt + x0 = ½a0t2 + v0t + x0 .

Example

A block, on a plane inclined at 30°, is initially moving up the plane with an initial velocity of 3m/s.  How far up the plane does the block go?

Neglect friction, decompose forces into components perpendicular to and parallel to the plane.  The parallel componenet is constant, and gives a constant acceleration.  Use the formulae for a constant acceleration, with an initial velocity, and find the time when the velocity becomes zero.  That is the time when the block is at its highest point.  Put that time into the equation for position and determine the maximum position up the plane.

Tension

In the third problem in the first homework assignment, two masses are connected by a rope.  If a rope (string, wire, chain, etc.) is under tension, it means that any piece in the rope is subject to equal and opposite forces along the direction of the rope.  At the ends, the rope pulls on whatever object it is attached to.

Summary of some other 217 Physics review topics