Problem 87 - Calculus for Biology and Medicine

Question

Neglecting air resistance, the height \(h\) (in meters) of an object thrown vertically from the ground with initial velocity \(v_{0}\) is given by $$ h(t)=v_{0} t-\frac{1}{2} g t^{2} $$ where \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) is the earth's gravitational constant and \(t\) is the time (in seconds) elapsed since the object was released. (a) Find the velocity and the acceleration of the object. (b) Find the time when the velocity is equal to \(0 .\) In which direction is the object traveling right before this time? in which direction right after this time?

Step-by-Step Solution

Answer

(a) Velocity: \(v(t) = v_0 - gt\); Acceleration: \(a(t) = -g\). (b) The velocity is zero at \(t = \frac{v_0}{g}\); upward before, downward after.

Step 1: Determine Velocity Function

The velocity of the object is the derivative of the height function with respect to time. If \(h(t) = v_0 t - \frac{1}{2} g t^2\), then the velocity \(v(t)\) is:\[v(t) = \frac{dh}{dt} = v_0 - gt\].

Step 2: Determine Acceleration Function

The acceleration of the object is the derivative of the velocity function with respect to time. Since \(v(t) = v_0 - gt\), the acceleration \(a(t)\) is:\[a(t) = \frac{dv}{dt} = -g\].This indicates that the acceleration is constant and equal to \(-9.81 \text{ m/s}^2\).

Step 3: Find Time When Velocity is Zero

To find when the velocity is zero, set the velocity function equal to zero and solve for \(t\):\[v(t) = v_0 - gt = 0\]\[v_0 = gt\]\[t = \frac{v_0}{g}\].

Step 4: Determine Direction Before and After Zero Velocity Time

Before \(t = \frac{v_0}{g}\), the velocity \(v(t) = v_0 - gt\) is positive, indicating that the object is moving upwards. After \(t = \frac{v_0}{g}\), the velocity becomes negative as \(v(t) = v_0 - gt\) becomes less than zero, indicating the object is moving downwards.

Key Concepts

Velocity

When studying the motion of an object under the influence of gravity, velocity is a crucial concept. Velocity tells us how fast and in which direction an object is moving. It is a vector quantity, meaning it has both magnitude and direction, like a car driving north at 60 km/h. For an object thrown vertically, like in our scenario, the velocity changes continuously because of gravity.
This change is represented by the velocity function, which is the derivative of the height function with respect to time. For our height function:

• The derivative or velocity function is given by:
  \[ v(t) = v_0 - gt \]
• Here, \( v_0 \) is the initial velocity and \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \))

In simpler terms, as time \( t \) increases, the term \( -gt \) becomes larger, causing the velocity to decrease as gravity pulls the object back down.

Acceleration

Acceleration is what describes how the velocity of an object changes over time. It is the rate of change of velocity and can be constant or variable. In our problem, since the object is under the sole influence of gravity, acceleration is constant. This constant acceleration due to gravity is also responsible for decreasing the velocity over time.

• In our scenario, the acceleration function is the derivative of the velocity function: \[ a(t) = \frac{dv}{dt} = -g \]
• This indicates that the acceleration is simply \( -9.81 \text{ m/s}^2 \) for the entire duration of the object's flight.

This acceleration acts downwards, hence the negative sign, showing that gravity decelerates an upward movement and accelerates a descent.

Free fall

The concept of free fall refers to the motion of an object where gravity is the only force acting upon it. When you drop a stone or throw a ball upwards, it is momentarily in free fall. During free fall, all objects experience the same acceleration irrespective of their masses, which is the gravitational acceleration \( g \) or \( 9.81 \text{ m/s}^2 \).

• In the example exercise, when the object is thrown, it rises until gravity slows it to a stop (velocity = 0), after which it falls back down.
• This ongoing change in motion solely due to gravity illustrates free fall.

Free fall is an essential concept in kinematics as it helps predict how long an object will stay in the air, how high it will go, and how fast it will be going when it hits the ground.

Differentiation

Differentiation is a math process used in kinematics and other areas of physics to find rates of change. It helps us understand how quantities like velocity and acceleration evolve over time. In the exercise provided, you encountered differentiation when you derived velocity and acceleration from the height function of an object.

• The velocity was found by differentiating the height with respect to time: \[ v(t) = \frac{dh}{dt} \]
• Then the acceleration was found by differentiating the velocity with respect to time: \[ a(t) = \frac{dv}{dt} \]

Differentiation simplifies complex motions into understandable functions. This allows us to predict and analyze the object’s behavior throughout its motion, making it a fundamental tool in understanding physical phenomena.