Applications of Differentiation – Applications of Derivatives: Position, Velocity, Acceleration

From our study of calculus so far, we know that the average rate of change is the slope of the secant line between two points on the graph. The instantaneous rate of change is the slope of the tangent line at a single point. Let us connect these points to position, velocity, and acceleration.

Position and Velocity

From your study of physics, you know to find the average velocity of a particle, you divide the total displacement (the straight-line distance between the final and initial position) by the total time:

Essentially, if we look at a position-time graph, the average velocity is the slope of the secant line between two points.

To obtain the instantaneous velocity at a particular time, we need the slope of the tangent line at the particular point on the position-time graph. In other words, we need to take the derivative of the position function P(t) to obtain the velocity function:

Velocity and Acceleration

From your study of physics, you know that to obtain the average acceleration, you divide the change in velocity by the total time:

Essentially, if we look at a velocity-time graph, the average acceleration is the slope of the secant line between two points.

To obtain the instantaneous acceleration at a particular time, we need the slope of the tangent line at the particular point on the velocity-time graph. In other words, we need to take the derivative of the Velocity Function V(t) to obtain the acceleration function:

Analyzing Motion Graphs

It’s also important to analyze the shapes of displacement graphs to approximate the shapes of velocity and acceleration graphs.

Let’s examine the position, velocity, and acceleration graph below:

Example 1:

Suppose we have the position function P(t) = t2 – 10t + 2. Then find the acceleration at any time t.

Solution:

As we know that if we are given the position and we have to find the acceleration, then we have to take the double derivative of the position.

So, by taking the second derivative of the position function,

P’(t) = V(t) = 2t – 1

P’’(t) = A(t) = 2

Therefore, the acceleration at any time t is 2.

Example 2:

Suppose we are given the position as P(t) = 2t3 + 4t -1. If the initial acceleration is zero, find the velocity of the particle.

Solution:

The first derivative of position will be velocity. So, we will take the derivation of P(t)

P’(t) = V(t) = 6t2 + 4

Taking the derivative of the velocity function, or taking the second derivative of the position function, we will get:

P’’(t) = V’(t) = 12t.

Because the acceleration is zero, set A(t) = 12t equal to zero and solve for t.

0 = 12t

t = 0

Now to find the value of velocity, put this value of ‘t’ into the velocity function V(t) = 6t2 + 4

= 6(0) + 4

= 4

So, velocity will be equal to four.

Example 3:

Suppose you have a function P(t) = 4t2 + 10t +10. This function gives the height of an object at time t. What is the velocity function? At what time is the velocity zero? What is the position of the article when the velocity is zero?

Solution:

Our position function is P(t) = 4t2 + 10t – t. Taking the derivative of P(t),

P’(t) = V(t) = 8t + 10

Set V(t) equal to zero to find the time when the velocity is zero:

0 = 8t + 10

t = 10 / 8

t = 1.25

The velocity is zero when t = 1.25 seconds.

Put t =1.25 inside P(t) to get the height of the object when the velocity is zero:

P(t) = 4(1.25)2 + 10(1.25) -1

P(t) = 6.25 +12.5 – 1

P(t) = 17.75

When the velocity is zero, the object is 17.75 units in the air.

Example 4:

A bug begins to crawl up a vertical wire at time t = 0. The velocity, V(t), of the bug at time t, 0 t 8 is given by the function whose graph is shown below.

1. At what value of t does the bug change direction? Justify your response.
2. During which time interval(s) is the bug slowing down? Justify your response.

Solution:

1. For position-time graphs, if the velocity changes from positive to negative (or negative to positive), then the object changes direction. The velocity is positive when the curve falls above the t-axis (object is going to the right). This occurs from t = 0 to t = 6. From t = 6 to t = 8, the curve falls below the t-axis, which means the velocity negative (object is going to the left). The transition from positive to negative occurs at t = 6. Therefore, the bug changes direction at t = 6.
2. The bug slows down, then the velocity and acceleration (slope of velocity curve) have opposite signs. From [1,2], both the velocity and acceleration are positive, so the bug is speeding up. From [2,4], velocity is positive, but acceleration is zero, which means the bug is going at a constant speed. From [4,6], the velocity is positive, but the acceleration is negative, so the bug is slowing down. From [6,7], both velocity and acceleration are negative, so the particle is speeding up. From [7,8], the velocity is negative, but the acceleration is positive, so the bug is slowing down. Therefore, the bug is slowing down on the time intervals of [4,6] and [7,8].

Example 5:

The figure graphed below shows the velocity of a particle moving along a coordinate line. Justify each response.

1. When is the particle moving right?
2. When is the particle moving left?
3. When is the particle stopped?
4. When is the particle speeding up?
5. When does the particle change directions?
6. When is the particle slowing down?
7. When is the particle moving at its greatest speed?
8. When is the particle’s acceleration positive?
9. When is the particle’s acceleration negative?

Solution:

1. The particle is moving right from t = 4 to t = 10. This is because the curve is above the t axis, which means the velocity is positive, which means that the particle is moving to the right.
2. The particle is moving left from t = 0 to t = 4. This is because the curve is below the t axis, which means the velocity is negative, which means that the particle is moving to the left.
3. The particle is stopped at t = 4. This is because this is the time when the curve crosses the t – axis, or when velocity is zero.
4. The particle is speeding up when both the acceleration (gradient of velocity curve) and velocity are positive (or when both are negative).
   1. This occurs at t = 0 to t = 1 (velocity and acceleration are negative).
   2. t = 4 to t = 5 (velocity and acceleration are positive)
   3. t = 7 to t = 9 (velocity and acceleration are positive)
5. The particle changes direction when the velocity changes from positive to negative and vice versa. This occurs when the curve crosses the t axis. This occurs at t = 4.
6. The particle slows down then the velocity decreases to zero (curve intercepts the t axis) where the velocity is zero.
   1. t = 3 to t = 4. The velocity is increasing to zero.
   2. t = 5 to t = 7. The velocity is decreasing to zero.
   3. t = 9 to t = 10. The velocity is decreasing to zero.
7. Speed is the absolute value of velocity. Therefore, we have to compare the highest point and lowest point on the graph. The highest point occurs at t = 5, v(t) = 4 m/s. The lowest point happens at t = 1 to t =3, v(t) = -4 m/s. Taking the absolute value of the velocity at the high and low points, they both equal 4 m/s. Therefore, the highest speed occurs at t = 1 to t = 3 and t = 5.
8. The acceleration is positive when the gradient of the velocity curve is positive. The gradient is positive from t = 4 to t = 5 and from t = 7 to t = 9.
9. The acceleration is negative when the gradient of the velocity curve is negative. The gradient is negative from t = 0 to t = 1, from t = 5 to t = 7, and from t = 9 to t = 10.

References:
https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the-derivative/distance-velocity-and-acceleration
http://www.pedersenscience.com/uploads/5/6/3/8/56384743/how_read_position_velocity_acceleration_graphs.pdf
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/motgraph.html