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 (s_n = x_n – x_n–1) traveled during the n^th second is given by

s_n = v₀ + a(n– ½)

[If you use the symbol u for the initial velocity, the above equation becomes s_n = 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 20^th 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

v_0x = v₀ cosθ₀ and

v_0y = 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

v_x = v₀ cosθ₀ and

v_y = v₀ sinθ₀ – gt

(7) Time of flight(T_f ) of the projectile is given by

T_f = (2 v₀ sinθ₀) /g

(8) Horizontal range (R) of the projectile is given by

R = (v₀²sin 2θ₀)/g

(i) The horizontal range is maximum (R_max) when the angle of projection θ₀ = 45º and

R_max = 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.