Problem 36 - Fundamentals of Physics

Question

A car moves along an \(x\) axis through a distance of \(900 \mathrm{~m}\), starting at rest (at \(x=0\) ) and cnding at rest (at \(x=900 \mathrm{~m}\) ). Through the first \(\frac{1}{4}\) of that distance, its acceleration is \(+2.25 \mathrm{~m} / \mathrm{s}^{2}\). Through the rest of that distance, its acceleration is \(-0.750 \mathrm{~m} / \mathrm{s}^{2}\). What are (a) its travel time through the \(900 \mathrm{~m}\) and \((\mathrm{b})\) its maximum speed? (c) Graph position \(x,\) velocity \(v,\) and acceleration \(a\) versus time \(t\) for the trip.

Step-by-Step Solution

Answer

(a) 56.55 s, (b) 31.81 m/s.

Step 1: Calculate the distance for the first phase

The problem tells us the car covers the first \(\frac{1}{4}\) of the total distance with an acceleration of \(2.25 \, \mathrm{m/s^2}\). Thus, the distance for this phase is \(\frac{1}{4} \times 900 \, \mathrm{m} = 225 \, \mathrm{m}\).

Step 2: Calculate time to cover 225 m with constant acceleration

The initial velocity \(v_0\) is 0, and the acceleration \(a\) is \(2.25 \, \mathrm{m/s^2}\). Using the equation \(s = v_0 t + \frac{1}{2} a t^2\), we can solve for time \(t\): \[225 = \frac{1}{2} \times 2.25 \times t^2\]\[t^2 = \frac{225}{1.125} = 200\]\[t = \sqrt{200} = 14.14 \, \mathrm{s}\].

Step 3: Determine the maximum velocity during first phase

Using the final velocity formula \(v = v_0 + a t\), where \(v_0 = 0\) and \(a = 2.25 \, \mathrm{m/s^2}\), we find:\[v = 2.25 \times 14.14 = 31.81 \, \mathrm{m/s}\].Thus, this is the maximum speed the car reaches before it starts decelerating.

Step 4: Calculate time to stop from maximum speed with negative acceleration

The remaining distance is \(900 \, \mathrm{m} - 225 \, \mathrm{m} = 675 \, \mathrm{m}\), and the deceleration \(a\) is \(-0.75 \, \mathrm{m/s^2}\). We use the equation \(v^2 = u^2 + 2 a s\), where \(v = 0\) and initial velocity \(u = 31.81 \, \mathrm{m/s}\), to find the distance during deceleration:\[0 = (31.81)^2 + 2(-0.75)s\]\[0 = 1011.56 - 1.5s\]\[s = \frac{1011.56}{1.5} = 675 \, \mathrm{m}\].This confirms the car stops once it covers the second segment of the distance.

Step 5: Calculate time for deceleration phase

With the maximum velocity \(u = 31.81 \, \mathrm{m/s}\) and acceleration \(a = -0.75 \, \mathrm{m/s^2}\), the time to stop is given by \(v = u + a t\) (where \(v=0\)):\[0 = 31.81 - 0.75t\]\[t = \frac{31.81}{0.75} = 42.41 \, \mathrm{s}\].

Step 6: Total travel time

The total travel time is the sum of the time taken in the first phase and the second phase:\[t_{total} = 14.14 \, \mathrm{s} + 42.41 \, \mathrm{s} = 56.55 \, \mathrm{s}\].

Key Concepts

Constant Acceleration

In the realm of kinematics, constant acceleration occurs when the acceleration of an object remains unchanged over a given period of time. This means the velocity of the object increases at a steady rate. For a car starting from rest, as in our exercise, this is an ideal situation to analyze because of the simplicity it provides in calculation. When acceleration is constant, you can use straightforward kinematic equations to predict distances traveled and velocities achieved.
With constant acceleration, like the car's initial phase with an acceleration of \(2.25 \, \mathrm{m/s^2}\), you can determine how far it moves using the equation \(s = v_0 t + \frac{1}{2} a t^2\). Since the car starts from rest, \(v_0\) is zero, simplifying calculations.
This concept is crucial for solving problems related to motion, and understanding it can help decipher many real-world situations where similar conditions apply. Constant acceleration helps in predicting how the motion evolves, be it in physics problems or engineering applications.

Maximum Velocity

Maximum velocity refers to the highest speed an object reaches during its motion under constant acceleration. In simpler terms, it's the peak speed before changing the motion parameters.
In the initial acceleration phase of our exercise, the car reaches its maximum velocity just before the deceleration phase begins. Here, maximum velocity is calculated using the formula \(v = v_0 + at\), where \(v_0\) is the initial speed (which is 0 for the car starting from rest) and \(a\) is the acceleration.
For the car, the calculated maximum velocity in the problem happens to be \(31.81 \, \mathrm{m/s}\). It's an important metric because it represents the vehicle's fastest speed during its journey, illustrating how quickly the car accelerates and the efficiency of its movement.

Deceleration

Deceleration is essentially negative acceleration, which happens when an object slows down. After reaching maximum velocity, the car begins decelerating. This happens in the remaining 675 meters of its journey.
When deceleration occurs, we can use the kinematic equations to determine time, velocity, or distance. In the second phase of the exercise, the car experiences a deceleration of \(-0.75 \, \mathrm{m/s^2}\). Using the equation \(v^2 = u^2 + 2as\), the problem confirms the car comes to a stop over the remaining distance.
Understanding deceleration is key to foreseeing how motion changes when forces oppose the direction of movement. This concept is applicable in numerous scenarios, such as braking in vehicles, and provides essential insights for safety and design in transportation.

Distance Traveled

The distance traveled by an object is the total length of its path, which in kinematics, is calculated using velocity and time under specified conditions. In our exercise, the total distance is broken into specific segments allowing for detailed calculations.
During the first phase, the car covers 225 meters as it speeds up, calculated by determining the portion of the journey completed at constant acceleration. Then, in the remaining phase, it travels 675 meters as it slows down to a stop.
Understanding and calculating distance traveled is vital in motion analysis. It informs us on how the change in velocity over time translates into spatial movement. It's an essential concept in designing travel routes, predicting travel time, and ensuring efficient energy use in vehicles.