Problem 45 - Calculus and its applications

Question

Given $$ s(t)=-10 t^{2}+2 t+5 $$ where \(s(t)\) is in meters and \(t\) is in seconds, find each of the following. a) \(v(t)\) b) \(a(t)\) c) The velocity and acceleration when \(t=1\) sec

Step-by-Step Solution

Answer

a) \( v(t) = -20t + 2 \); b) \( a(t) = -20 \); c) Velocity: \( -18 \) m/s, Acceleration: \( -20 \) m/s² at \( t=1 \) sec.

Step 1: Differentiate to Find Velocity

To find the velocity function, differentiate the position function \( s(t) = -10t^2 + 2t + 5 \) with respect to time \( t \). The derivative gives the velocity function \( v(t) \).The derivative is: \[ v(t) = \frac{d}{dt}[-10t^2 + 2t + 5] = -20t + 2 \] Thus, the velocity function is \( v(t) = -20t + 2 \).

Step 2: Differentiate Again to Find Acceleration

To find the acceleration function, differentiate the velocity function \( v(t) = -20t + 2 \) with respect to time \( t \). The derivative gives the acceleration function \( a(t) \).The derivative is: \[ a(t) = \frac{d}{dt}[-20t + 2] = -20 \] Thus, the acceleration function is \( a(t) = -20 \), which is constant.

Step 3: Evaluate Velocity at \( t=1 \) Second

Substitute \( t = 1 \) into the velocity function \( v(t) = -20t + 2 \) to find the velocity at \( t = 1 \) second.Calculate: \[ v(1) = -20(1) + 2 = -20 + 2 = -18 \] The velocity when \( t = 1 \) second is \( -18 \) meters per second.

Step 4: Evaluate Acceleration at \( t=1 \) Second

Since the acceleration is constant \( a(t) = -20 \), it does not depend on \( t \). Therefore, at \( t = 1 \) second, the acceleration is still \( -20 \) meters per second squared.

Key Concepts

Velocity function

The velocity function, denoted as \( v(t) \), represents the rate of change of position with respect to time. Think of velocity as the speed at which an object is moving along with its direction. To find the velocity function from the given position function \( s(t) = -10t^2 + 2t + 5 \), we must differentiate it with respect to time \( t \). Differentiation helps us determine how the position changes at any instant.

For the function \( s(t) = -10t^2 + 2t + 5 \), the derivative with respect to \( t \) is:

• Differentiate \( -10t^2 \) to get \(-20t\).
• Differentiate \( 2t \) to get \(+2\).
• Constant \( 5 \) differentiates to \( 0 \).

Therefore, the velocity function is \( v(t) = -20t + 2 \). This means at any time \( t \), the velocity is computed by plugging \( t \) into this function.

Understanding the velocity function can tell you how quickly the object is moving and in what direction at any point in time.

Acceleration function

The acceleration function, represented as \( a(t) \), describes the rate of change of velocity with respect to time. It's essentially how quickly the velocity is changing. To find \( a(t) \), we differentiate the velocity function \( v(t) = -20t + 2 \) with respect to \( t \).

Differentiating \( v(t) \):

• The derivative of \(-20t\) is \(-20\).
• The derivative of the constant \(+2\) is \(0\).

Thus, the acceleration function is \( a(t) = -20 \). This shows that the acceleration is constant and negative, implying that the object is decelerating or slowing down at a steady rate.

A constant acceleration makes it easy to predict motion since changes in velocity occur at a uniform rate.

Position function

The position function \( s(t) \) describes where an object is located at any given time \( t \). In the exercise, the position function is given as \( s(t) = -10t^2 + 2t + 5 \). This function provides crucial information about the trajectory of the object, revealing patterns such as initial position and how the position changes over time.

This quadratic function implies that the object is moving along a parabolic path. It accounts for:

• The term \(-10t^2\) suggests that the position changes quadratically, influencing the curvature of the motion.
• The term \(2t\) indicates a linear component impacting velocity directly.
• The constant \(5\) represents the initial position of the object at \( t = 0 \).

By analyzing \( s(t) \), we can gain insights into the object's journey, including initial setup and how quickly it reaches different points along its path.

Knowing the position function is foundational as it sets the stage for understanding both velocity and acceleration over time.