Kinematic Equations
What five variables are used in the three kinematic equations?
The kinematic equations use ∆x, v, a, t, and v0.
(displacement, velocity, acceleration, time, initial velocity)
What do vectors have that scalars lack?
Vectors have direction.
(Both have magnitude, but scalars have no direction.)
What is the value of acceleration if velocity is constant?
If velocity is constant, acceleration is zero.
(Constant velocity means unchanging velocity, and acceleration is change in velocity.)
What are the three kinematic equations for the y direction?
vy = v0y + ayt
∆y = v0yt + 1/2 ayt2
vy2 = v0y2 + 2ay∆y
(Add y as a subscript to velocities and acceleration and change ∆x to ∆y.)
For a vector that has a horizontal component of 6.00 km and a direction of 75° above the horizontal, what are the magnitude and the vertical component?
Magnitude = 23.2 km
Vertical = 22.4 km
(cos θ = horizontal / magnitude
magnitude = horizontal / cos θ;
tan θ = vertical / horizontal
vertical = horizontal · tan θ)
A ball is dropped from rest from a 20.0 m cliff. How long does it take to land, and how fast is it moving on impact? Assume a constant downward acceleration of 9.80 m/s².
Time = 2.02 s
Final velocity = 19.8 m/s
(Δy = 1/2 at² (position equation, no initial velocity)
t = √(2 · ∆y / a).
v = at (velocity equation, no initial velocity))
If an object starting from rest and undergoing constant acceleration a goes distance d in time t, how far (compared to d) does it go in time 2t?
4d
(∆x = 1/2 at2
d = 1/2 at2
Let D be the distance it goes in time 2t.
D = 1/2 a(2t)2 = 1/2 a(4t2) = 4(1/2 at2) = 4d)
An object undergoes a displacement of 3.0 m at 60° above the horizontal, then a displacement of 1.1 m at 105° above the horizontal. What are the magnitude and direction of the final displacement?
Magnitude = 3.9 m
Direction = 72° above horizontal
(Displacement 1: 3.0 cos 60° m right, 3.0 sin 60° m up
Displacement 2: 1.1 cos 105° m right, 1.1 sin 105° m up
Total displacement: 3.0 cos 60° + 1.1 cos 105° m right, 3.0 sin 60° + 1.1 sin 105° m up = 1.2 m right, 3.7 m up
Final magnitude = √(1.22 + 3.72) m = 3.9 m
Final angle = tan-1(3.7 / 1.2) = 72° above horizontal)
A projectile is launched from ground level at an angle of 45° above the horizontal and a speed of 100. m/s. How far away does it land, i.e., what is the horizontal component of its displacement? Assume constant downward acceleration of 9.80 m/s2.
1020 m
(It lands at the same elevation so ∆y = 0.
0 = ∆y = v0yt + 1/2 ayt2 = t(v0y + 1/2 ayt)
t = -2v0y / ay
∆x = v0xt + 1/2 axt = v0xt = -2v0xv0y / ay
Find v0x and v0y using the initial speed and angle and plug in.)