v = u + at
(velocity-time equation)s = ut + 1/2at^2
(displacement-time equation)v^2 = u^2 + 2as
(velocity-displacement equation)s = vt - 1/2at^2
(displacement-velocity equation)average speed = total distance / time
average velocity = displacement / time
An object’s position x, in meters, obeys the equation x = sin(t), where t is the time in seconds since the object began moving. How fast is the object moving at t = π/2 seconds?
- Solution
- This question asks for the velocity of the object at t = π/2 seconds.
- Since velocity is the time derivative of position, the velocity is given by
v = dsin(t) = cos(t)
.- Then plug in the value of t:
v(π/2) = cos(π/2) = 0 m/s
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Acceleration measures the rate-ofchange of an object’s velocity. An object’s average acceleration is defined as follows:
Acceleration Sample Problem
A car is traveling along a straight highway at a speed of 20 m/s. The driver steps on the gas pedal and, 3 seconds later, the car’s speed is 32 m/s. Find its average acceleration.
- Solution
- Here the direction does not change but the speed does, so there is a change in velocity.
- To obtain the average acceleration, simply divide the change in velocity,
32 m/s – 20 m/s = 12 m/s
, by the time interval during which the change occurred:= ∆v/∆t = (12 m/s)/(3 s) = 4 m/s 2
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Projectile motion is the motion of an object that is launched into the air and then moves under the influence of gravity.
An object is projected upward with a 30° launch angle and an initial speed of 60 m/s. For how many seconds will it be in the air? How far will it travel horizontally before returning to its original height?
- Solution
- The total time the object spends in the air is equal to twice the time required to reach the top of the trajectory (because the parabola is symmetrical). So, as we did in the previous example, we find the time required to reach the top by setting vy equal to 0, and now double that amount of time:

Therefore, the total flight time (that is, up and down) is T = 2t = 2 × (3 s) = 6 s.
An object is projected upward with a 30° launch angle and an initial speed of 40 m/s. How long will it take for the object to reach the top of its trajectory? How high is this?
- Solution
- When the projectile reaches the top of its trajectory, its velocity vector is momentarily horizontal; that is, vy = 0. We are asked how long it will take the object to reach this point with θ = 30 degrees. Using the vertical-motion equation for vy , we can set it equal to 0 and solve for t:

At this time, the projectile’s vertical displacement is
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A rock is dropped from an 80-meter cliff. How long does it take to reach the ground?
- Solution
- Since all of the rock’s motion is down, we call down the positive direction, so
a = +g
. We’re given v0 , ∆y, and a, and we are asked for t. So v is missing; it isn’t given and it isn’t asked for, and we use Big Five #3:

Consider an object moving along a straight axis in such a way that its velocity, v, as a function of time, t, is given by the following velocity-vs.-time graph:
At time t = 0, the object’s velocity was v = 0. Over the first two seconds, its velocity increased steadily to 10 m/s. At time t = 2 s, the velocity then began to decrease (eventually becoming v = 0, at time t = 3 s). The velocity then became negative after t = 3 s, reaching v = –5 m/s at time t = 3.5 s. From t = 3.5 s on, the velocity remained a steady –5 m/s.
First, the fact that the velocity changed from t = 0 to t = 2 s tells us that the object accelerated. The acceleration during this time was
Note, however, that the ratio that defines the acceleration, ∆v/∆t, also defines the slope of the v-vs.-t graph. Therefore, “The slope of a velocity-vs.-time graph gives the average acceleration.“
What was the acceleration from time t = 2 s to time t = 3.5 s? The slope of the line segment joining the point (t, v) = (2 s, 10 m/s) to the point (t, v) = (3.5 s, –5 m/s) is
What is the object’s acceleration from time t = 3.5 s to time t = 6 s? Since the line segment from t = 3.5 s to t = 6 s is horizontal, its slope is zero, which indicates that the acceleration is zero, but you can also see this from looking at the graph; the object’s velocity did not change during this time interval.
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v = u + at
: This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and time (t).s = ut + 1/2at^2
: This equation relates the displacement (s) of an object to its initial velocity (u), acceleration (a), and time (t).v^2 = u^2 + 2as
: This equation relates the final velocity (v) of an object to its initial velocity (u), acceleration (a), and displacement (s).Equations | Missing Variable | |
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Big Five #1 | ![]() | a |
Big Five #2 | ![]() | ∆x |
Big Five #3 | ![]() | v |
Big Five #4 | ![]() | v0 |
Big Five #5 | ![]() | t |
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