Question

The position of an object as a function of time is given as \(x=A t^{3}+B t^{2}+C t+D .\) The constants are \(A=2.10 \mathrm{~m} / \mathrm{s}^{3}, B=1.00 \mathrm{~m} / \mathrm{s}^{2}\), \(C=-4.10 \mathrm{~m} / \mathrm{s},\) and \(D=3.00 \mathrm{~m}\). a) What is the velocity of the object at \(t=10.0 \mathrm{~s} ?\) b) At what time(s) is the object at rest? c) What is the acceleration of the object at \(t=0.50\) s? d) Plot the acceleration as a function of time for the time interval from \(t=-10.0 \mathrm{~s}\) to \(t=10.0 \mathrm{~s}\).

Short answer

A: The object is at rest at \(t \approx 1.12s\).

Step-by-step solution

Velocity function and velocity at \(t=10.0s\)

To find the velocity function, we need to find the derivative of the position function with respect to time. The position function is given by: \(x(t)=2.10t^3+1.00t^2-4.10t+3.00\). The derivative will give us the velocity function: \(v(t) = \frac{dx}{dt} = \frac{d(2.10t^3+1.00t^2-4.10t+3.00)}{dt}\). Calculate the derivative: \(v(t) = 6.3t^2+2.0t-4.1\). Now plug \(t=10.0s\) into this velocity function to find the velocity at \(t=10.0s\): \(v(10) = 6.3(10)^2+2.0(10)-4.1 = 630+20-4.1 = 645.9 \frac{m}{s}\) b)

Finding the time when the object is at rest

The object is at rest when its velocity is zero. Use the velocity function we found earlier: \(v(t) = 6.3t^2+2.0t-4.1\). To find the time(s) when the object is at rest, we need to solve the equation \(v(t) = 0.\) Solving for t, we get two possible times: \(t \approx -0.59s\) and \(t \approx 1.12s\). However, the negative value for time is not physically meaningful, so we consider the object to be at rest at \(t \approx 1.12s\). c)

Acceleration function and acceleration at \(t=0.50s\)

The acceleration function is the derivative of the velocity function with respect to time. The velocity function we found earlier is \(v(t) = 6.3t^2+2.0t-4.1\). The derivative will give us the acceleration function: \(a(t) = \frac{dv}{dt} = \frac{d(6.3t^2+2.0t-4.1)}{dt}\). Calculate the derivative: \(a(t) = 12.6t+2.0\). Now plug \(t=0.50s\) into this acceleration function to find the acceleration at \(t=0.50s\): \(a(0.50) = 12.6(0.50)+2.0 = 6.3+2 = 8.3 \frac{m}{s^2}\) d)

Plotting the acceleration function

To plot the acceleration function for the time interval from \(t=-10.0s\) to \(t=10.0s\), we need to use the acceleration function we found earlier: \(a(t) = 12.6t+2.0\). Plot this function in the given time interval. You should use graphical software or online tools like Desmos, Geogebra, or another plotting tool to create this plot. The plot will be a straight line with a positive slope, passing through the point \((0,2)\) on the y-axis.

Key concepts

Kinematic Equations

Kinematic equations are the foundation of predicting an object's motion when the acceleration is constant. In our exercise, however, the acceleration isn't constant, highlighting a more advanced aspect of motion – time-dependent acceleration. The position function provided,
\(x(t)=At^3+Bt^2+Ct+D\),
represents the motion with changing acceleration. The kinematic equations simplify to this function when acceleration varies over time. By taking derivatives, we obtain the velocity and acceleration functions, connecting position, velocity, and acceleration mathematically. The understanding of these equations is crucial for solving problems related to motion, like the one presented.

Velocity and Acceleration

Velocity and acceleration are two core concepts in motion physics. Velocity is the rate of change of position over time and is a vector quantity, meaning it has both magnitude and direction. Acceleration, on the other hand, is the rate of change of velocity over time. In the given problem, we differentiate the position function to find the velocity:
\(v(t) = \frac{dx}{dt}\),
and then the velocity function to get the acceleration:
\(a(t) = \frac{dv}{dt}\).
These steps reveal how the object's speed and acceleration change throughout its journey. Using calculus allows us to handle non-uniform acceleration, a common scenario in real-world motions.

Time-Dependent Position Functions

Time-dependent position functions describe how an object's position changes over time. In this case, the provided cubic equation
\(x(t)=At^3+Bt^2+Ct+D\)
corresponds to a time-dependent position function where acceleration is not uniform. It's important to handle each term in the equation with care as each represents a different aspect of motion. The cubic term \(At^3\) indicates changing acceleration, while the linear term \(Ct\) would suggest constant acceleration if it existed on its own. By mastering time-dependent position functions, one can solve complex motion problems that involve varying forces and accelerations, similar to the ones encountered in real-life physics scenarios.