Question

The position of a particle of mass \(2.17\) kg traveling in a straight line is given by $$ x=\left(0.179 \mathrm{~m} / \mathrm{s}^{4}\right) t^{4}-\left(2.08 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}+17.1 \mathrm{~m} $$ Find the \((a)\) velocity, \((b)\) acceleration, and \((c)\) force on the particle at time \(t=7.18 \mathrm{~s}\)

Short answer

The velocity, acceleration, and force acting on the particle at \(t = 7.18 \, s\) can be found by calculating the first and second derivatives of the given position equation and then multiplying the mass of the particle with the acceleration to obtain the force. Detailed calculations are needed to reach the specific numerical results.

Step-by-step solution

Compute the Velocity

Velocity is the derivative of the position \(x\) with respect to time \(t\). We have to differentiate the given equation for position: \[x= (0.179 \, m / s^4) t^4 - (2.08 \, m / s^2) t^2 + 17.1 \, m\]. This gives us the velocity function \(v(t)\). Using the power rule for derivatives, the result is: \[v(t) = 4 \cdot (0.179 \, m / s^4) t^3 - 2 \cdot (2.08 \, m / s^2) t\].

Calculate the Velocity at t = 7.18 Seconds

Put \(t=7.18 \, s\) into the velocity equation from step 1 to get the velocity at that time point.

Compute the Acceleration

Acceleration is the derivative of velocity \(v(t)\) with respect to time \(t\). Now take the derivative of the velocity function we computed in step 1 to find the acceleration function \(a(t)\). This results in \[a(t) = 12 \cdot (0.179 \, m / s^4) t^2 - 2 \cdot (2.08 \, m / s^2)\].

Calculate the Acceleration at t = 7.18 Seconds

Just like in step 2, we now plug in \(t=7.18 \, s\) into the acceleration equation from step 3 to achieve the acceleration at that particular time.

Compute the Force

According to Newton's second law of motion, force is defined as mass times acceleration. Hence, we multiply the mass of the particle, \(2.17 \, kg\), by the acceleration we computed in step 4 to get the force.

Calculate the Force at t = 7.18 Seconds

We calculate the force acting on the particle at \(t = 7.18 \, s\) with the force equation from step 5.

Key concepts

Velocity

Understanding what velocity is and how to calculate it is fundamental in physics. Velocity tells us how fast an object is moving and in which direction. It is a vector quantity, which means it has both magnitude (speed) and direction. The magnitude is the speed of the object, while the direction is the path that the object is moving along.

In the given exercise, the velocity is found by differentiating the position function with respect to time. Utilizing your knowledge in differential calculus, apply the power rule to find the velocity equation. Essentially, for a position given by a function of time, the derivative of this function at any point gives you the instantaneous velocity at that time. Here, upon deriving the function, the calculated velocity at time 7.18 seconds offers a snapshot of how fast the particle was moving at that specific moment.

Acceleration

Acceleration is a measure of how quickly the velocity of an object is changing. It can be a result of an increase or decrease in speed or changing direction. It's also a vector quantity with both magnitude and direction. To find the acceleration of an object, we need to look at the rate at which the velocity is changing over time.

In the exercise, acceleration is derived from the velocity equation, which is itself derived from the position equation. By applying differential calculus a second time, acceleration, which is the derivative of velocity, reveals how the speed of our particle is changing at any given moment. Computing this value at time 7.18 seconds gives us the exact rate of this change at that specific point in time.

Force

Force can be considered as any interaction that causes an object to be pushed, pulled, or to experience a change in motion. In physics, a force can cause an object with mass to change its velocity, which includes starting from rest, increasing speed, decreasing speed, or changing direction. The unit of force in the International System of Units (SI) is the newton (N).

According to Newton's second law of motion, force is the product of mass and acceleration (\(F = ma\)). This formula is used in the exercise to calculate the force acting on the particle by multiplying the particle's mass with the acceleration found previously. When we find the force at time 7.18 seconds, we are determining the instantaneous force acting on the particle at that point in time.

Differential Calculus in Physics

Differential calculus is a branch of mathematics that deals with rates of change and slopes of curves. In the context of physics, it's particularly valuable as it allows us to calculate velocity and acceleration, which are all about rates of change—of position and velocity, respectively.

Through this exercise, it's illustrated that differential calculus is not just abstract mathematics, but a practical tool for solving real-world problems. By taking derivatives of position with respect to time, you can find velocity, and by taking derivatives of velocity, you can find acceleration. This demonstrates just how intertwined physics and calculus are when describing motion.

Newton's Second Law of Motion

Newton's second law of motion is one of the most significant principles in classical mechanics. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (\(F = ma\)). This law highlights the direct proportionality between force and acceleration, and additionally connects these with the mass of the object being accelerated.

When solving the problem in the exercise, we use Newton's second law to find the force on a particle. This law is foundational to understanding why the particle moves the way it does under certain forces. Applying this law allows students to predict the quantitative effect of forces on the motion of objects.