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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Wednesday, August 13, 2008

### Kinematics for AP Physics B & C- One Dimensional and Two Dimensional Motions

You will have to remember certain basic formulae for solving multiple choice questions within the permitted time. The essential things you need to remember in one dimensional uniformly accelerated motion are here:
(1) The final velocity (v) at time t of an object in one dimensional motion with uniform acceleration a is given by
v = v0 + at where v0 is the initial velocity (at time t = 0)
[If you use the symbol u for the initial velocity, the equation becomes v = u + at]
(2) The position x of the object at time t is related its initial position x0 (at zero time) as
x = x0 + v0 t+ (½) at2
This can be rewritten in terms of the displacement s = x x0 as
s = v0 t+ (½) at2
[If you use the symbol u for the initial velocity, the above equation becomes s = ut+ (½) at2]
(3) The final velocity is related to the displacement (x x0) as
v2 = v02 + 2a(x x0)
[If you use the symbol u for the initial velocity and the symbol s for the displacement, the above equation becomes v2 = u2 + 2as]
(4) The distance (sn = xn xn1) traveled during the nth second is given by
sn = v0 + a(n– ½)
[If you use the symbol u for the initial velocity, the above equation becomes sn = u + a(n– ½)]
(5) (i) The slope of the displacement – time graph (obtained by plotting time on the X-axis and the displacement on the Y-axis) gives the velocity.
(ii) The slope of the velocity – time graph (obtained by plotting time on the X-axis and the velocity on the Y-axis) gives the acceleration.
(iii) The area under the velocity – time graph (obtained by plotting time on the X-axis and the velocity on the Y-axis) gives the displacement.
In two dimensional motion you have to consider mainly circular motion and projectile motion. We have already discussed the essential points required in the case of circular motion in the post dated 20th January 2008. You can access that post as well as related posts by clicking on the label ‘circular motion’ below this post. Here are the important equations you need to remember in the case of projectile motion:
(6) In projectile motion, the gravitational force affects the vertical component of the velocity (of the projectile) only. If the projectile is launched with velocity v0 from the origin, making an angle θ0 with respect to the horizontal, the x-component and the y-component of the velocity v0 are respectively
v0x = v0 cosθ0 and
v0y = v0 sinθ0
The x and y co-ordinates of the projectile after time t are respectively
x = (v0 cosθ0)t and
y = (v0 sinθ0)t – (½)gt2
The x and y components of the velocity of the projectile after time t are respectively
vx = v0 cosθ0 and
vy = v0 sinθ0 gt
(7) Time of flight(Tf ) of the projectile is given by
Tf = (2 v0 sinθ0) /g
(8) Horizontal range (R) of the projectile is given by
R = (v02sin 2θ0)/g
(i) The horizontal range is maximum (Rmax) when the angle of projection θ0 = 45º and
Rmax = v02/ g
(ii) If the velocity of projection v0 is the same, the horizontal range is the same for angles of projection θ0 and (90º θ0). These directions are equally inclined to the 45º direction for maximum range.
(10) Maximum height (H) reached by the projectile is given by
H = (v02sin2θ0)/2g
If the projectile is launched vertically, the maximum height reached will be v02/ 2g, which is half the maximum range.
(11) for any given value of the angle of projection θ0, the horizontal range R and the maximum height H are related as
R = 4H cot θ0
We will discuss questions from this section in the next post. Meanwhile, you may go through some useful multiple choice questions on one dimensional motion given here.
You can find useful posts on two dimensional motion here.