Problem 105 - Fundamentals of Physics

Question

A motorcycle is moving at \(30 \mathrm{~m} / \mathrm{s}\) when the rider applies the brakes, giving the motorcycle a constant deceleration. During the \(3.0 \mathrm{~s}\) interval immediately after braking begins, the speed decreases to \(15 \mathrm{~m} / \mathrm{s} .\) What distance does the motorcycle travel from the instant braking begins until the motorcycle stops?

Step-by-Step Solution

Answer

The motorcycle travels 90 meters before stopping.

Step 1: Determine the deceleration

The initial speed of the motorcycle, \( v_i \), is \( 30 \ \mathrm{m/s} \) and the final speed after \( 3 \ \mathrm{s} \), \( v_f \), is \( 15 \ \mathrm{m/s} \). The time, \( t \), is \( 3 \ \mathrm{s} \). Use the formula for acceleration: \( a = \frac{v_f - v_i}{t} \).\[ a = \frac{15 \ \mathrm{m/s} - 30 \ \mathrm{m/s}}{3 \ \mathrm{s}} = -5 \ \mathrm{m/s}^2 \] The negative sign indicates deceleration.

Step 2: Use deceleration to find stopping distance

Now we need to find the total distance traveled until the motorcycle stops completely. Set the final velocity \( v_f \) to 0 \( \mathrm{m/s} \) when the motorcycle stops. Use the formula:\[ v_f^2 = v_i^2 + 2a d \] Solving for \( d \), the stopping distance:\[ 0 = 30^2 + 2(-5) d \] \[ 0 = 900 - 10d \] \[ 10d = 900 \] \[ d = 90 \ \mathrm{m} \] Thus, the motorcycle travels a total distance of 90 meters before coming to a stop.

Key Concepts

Deceleration

Deceleration occurs when an object is slowing down, or when there's a reduction in speed. It's similar to acceleration but in the opposite direction. In physics, it's important because it affects how fast something comes to a stop.
Deceleration is often represented by a negative acceleration because it reduces the velocity of a moving object. When solving problems involving deceleration, we use the formula for acceleration, which is given by:

• \( a = \frac{v_f - v_i}{t} \)

Here, \( v_i \) and \( v_f \) are the initial and final velocities, respectively, and \( t \) is the time it takes for the change in velocity. A negative result from this formula confirms deceleration.
Understanding deceleration is crucial for solving many physics problems, such as those involving vehicles coming to a halt.

Stopping Distance

Stopping distance is the total distance an object covers from the point where brakes are applied until it fully stops. In our motorcycle example, it is crucial to calculate how far the motorcycle travels before it comes to a standstill.
To find the stopping distance, we use the equation:

• \( v_f^2 = v_i^2 + 2a d \)

Here, \( v_f \) is the final velocity (which is zero when the object stops), \( v_i \) is the initial velocity, \( a \) is the acceleration (negative for deceleration), and \( d \) is the stopping distance we need to find.
By rearranging and solving this equation, one can determine how long it takes for a vehicle to reach a full stop after the brakes are applied.

Kinematics

Kinematics is a branch of classical mechanics that focuses on the motion of objects without considering the forces that cause these motions. It involves parameters like displacement, velocity, and acceleration.
In kinematic problems, equations are used to relate these parameters to understand how an object moves. The key equations include those for calculating acceleration, stopping distance, and final velocity.

• For uniform acceleration or deceleration, the following equations are used:
• \( v = v_i + a \, t \)
• \( s = v_i \, t + 0.5 \, a \, t^2 \)
• \( v^2 = v_i^2 + 2 \, a \, s \)

Mastering kinematics is beneficial for efficiently solving problems involving linear motion, as seen with the motorcycle example.

Constant Acceleration

Constant acceleration means the rate of change of velocity with respect to time stays the same throughout the motion. In other words, acceleration doesn’t increase or decrease as the object moves, making calculations more straightforward.
In the context of our problem, once the brake is applied to the motorcycle, it experiences a constant deceleration of \(-5 \mathrm{m/s}^2\). This makes it simple to use kinematic equations directly, as they assume constant acceleration.
Using the formula \( a = \frac{v_f - v_i}{t} \), we can determine the acceleration or deceleration during the motion phase. Understanding this concept is essential for evaluating motion scenarios effectively. These principles play a vital role in accurately predicting the motion path and required stopping distance.