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Kinematics – describes the motion of object without causes that leaded to the motion We are not interested in details of the object (it can be car, person, box etc..). We treat it as a point We want to describe position of the object with respect to time – we want to know position at any given time Path (trajectory) – imaginary line along which the object moves

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x t

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Motion along a straight line We will always try to set up our reference frame in a such way that motion is along or “x” or “y” coordinate axis. The direction of the axis is up to us. Position vector then can be represented by a single component, the other components are equal to zero. r(t 1 ) r(t 2 ) r(t 3 ) x y t=t 1 t=t 2 t=t 3 x 1 =19m,t 1 =1s x 2 =277m,t 1 =4s Very often I will write r n instead of r(t n ), the same for the components of the vector, for example, y n instead of y(t n )

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Motion along a straight line x component of a displacement vector – for time interval t 1 t 2 is equal D x=x 2 -x 1 Note that we can define another reference frame, position vector will be different in each frame, not a displacement vector (it will have different components) r(t 1 ) r(t 2 ) x y r’ 1 r’ 2 x’ y’

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So we can define change in time t=t 2 -t 1 – or in other words a time interval during which x occurs. We will define x-component of an object’s average velocity = x/ t=(x 2 -x 1 )/(t 2 -t 1 )=v av. X Units - m/s X-component of the velocity is equal to a slope of a line in x-t plane that passes through points (x 1,t 1 ) and (x 2,t 2 ) X (m) t (sec) DtDt DxDx

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The value of x-component of an average velocity is defined for given interval. Note, since x-component of the displacement can be positive (motion in +x) or negative (motion in -x), x-component of an average velocity can be positive or negative. Note that if you start at x 1 =x 2 then x-component of average velocity for given time interval is equal to zero. Average speed=distance traveled / time interval – positive quantity X (m) t (sec) Two time intervals Motion in positive x direction Motion in negative x-direction Distance is equal to sum of both

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Example: Figure shows the position of a moving object as a function of time. (a) find the average velocity of this object from points A to B, B to C and A to C. (b) is the average speed for intervals given in (a) will be less than, equal, or greater than the values found in part (a). x(m) t (s) A B C

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Example: Each graph in figure shows the position of a running cat called Mousie, as a function of time. In each case, sketch a clear qualitative (no numbers) graph of Mousie’s velocity as a function of time. X(m) t(s)

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x t Case we have more curves that describes motion several objects Points were graphs cross each other – place and time where both meet each other

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Instantaneous velocity The same way we can define an instantaneous speed. Note: we can always make the time interval small enough so distance traveled during Dt is equal to |Dx| (motion along straight line). Instantaneous speed is a magnitude of instantaneous velocity Average velocity as everything that is average does not give us chance to find out how was the object moving at any given time of the time interval. Instead we define (x-component) an instantaneous velocity, decreasing time interval.

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t1t1 t2t2 X (m) t (sec) We see that instantaneous velocity is slope of a tangent at point t 1

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X (m) t (sec) Positive slope Slope decreasing – slowing down, since motion in positive x Slope is zero – instant stop Slope is increasing – speeding up, since motion in positive x Zero slope - stopped Slope is decreasing, motion in negative direction, object speeding up

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Average acceleration Suppose that I have a particle that was moving somehow and I know that x-component of instantaneous velocity is given as v x (t). We want to know how much it changed during time interval D t=t 2 - t 1, D v = v x 2 - v x 1 X-component of an average acceleration of the particle is equal to a av x = D v/ D t UNITS: meters/sec 2 =m/s 2 We are getting average acceleration from instantaneous velocity, remember it is a vector. In our case we are considering only x- component of it. And can be positive or negative

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t1t1 t2t2 v (m/s) t (sec) DtDt D v x =v x2 -v x1 x-component of average acceleration in time interval D t is a slope of the straight line that passes through 2 points of v x vs t curve

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t1t1 t2t2 v (m/s) t (sec) We see that instantaneous acceleration is slope of a tangent to v x (t) curveat point t 1 Reducing time interval we can define an instantaneous acceleration

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Special cases Object is not moving: Object is moving with constant velocity – means slope to x(t) curve is constant – means it is again a straight line: Motion with constant velocity differs from motion with constant speed. The words constant velocity locks magnitude and direction – means object is moving along straight line x t x0x0 x t x0x0 vxvx t v x0 vxvx t

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x t x0x0 vxvx t t1t1 t2t2 x1x1 x2x2 t1t1 t2t2 Area of shaded rectangular region is equal to v x0 (t 2 -t 1 )=v x0 D t = D x x 2 -x 1 =D x=v av-x D t=v x0 D t Area of a region created by v x (t) curve, t=t 1, t=t 2 and v x =0 is equal to x- component of displacement.

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Motion with constant acceleration Instantaneous and average acceleration are equal x-component of velocity is given: v x (t)=v 0x +a x t One can see that at time t=0 v x =v 0x We know that displacement is equal to area bound by velocity versus time, t=t 1, t=t 2 and v x =0 lines – trapezoid. Height =h= t 2 -t 1 two parallel sides: side 1 =v 0x +a t 1 and side 2 =v 0x +a t 2 Expression for x-component of the position vector, when x 0, t, v 0x, a x are given Expression for x-components, when x 0,t, v 0x, v x are given (acceleration is not given) We can set t 1 =0 and write x 1 as x 0, indicating that position at t=0 Expression for x-components, when x 2 –x 1, v 1x, v 2x,,a x are given. (time is not given)

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Example: According to recent data, a Ford Focus travels 0.25mi in 19.9s, starting from rest. The same car, when braking from 60mph on dry pavement, stops in 146ft. (a) Find this car’s acceleration and deceleration, (b) assuming constant acceleration find its velocity after 0.25mi.

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Freely falling objects In this section we will consider constant acceleration due to gravity. Magnitude of the acceleration we will denote as g=9.8m/s 2 (for Earth) Acceleration due to gravity is a vector pointing toward the center of the (in our case Earth) we can choose our reference frame with y -axis pointing up – then y - component of the acceleration is equal to –g. We can use all equations for constant acceleration with a y =-g. Special case when you throw object upward, how high does it get?

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Example: If we throw a ball upward with initial velocity v 0y =5m/s, find: how high does it go, how long does it take to get there velocity of the ball when it is on its way back time required to return back Example: Measuring depth of a deep well To measure the depth of the well you drop a rock and start your stopwatch. Find the depth of the well if after 3s you hear hitting sound coming from a bottom of the well. Ignore finite speed of a sound.

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Example: Two rockets start from rest and accelerate to the same final speed, but one has twice the acceleration of the other. (a) If the high- acceleration rocket takes 50s to reach the final speed, how long will it take the other rocket to reach that speed. (b) If the high acceleration rocket travels 250m to reach its final speed, how far will the other rocket travel to do likewise?

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