Question

A runner of mass 57.5 kg starts from rest and accelerates with a constant acceleration of \(1.25 \mathrm{~m} / \mathrm{s}^{2}\) until she reaches a velocity of \(6.3 \mathrm{~m} / \mathrm{s}\). She then continues running with this constant velocity. a) How far has she run after 59.7 s? b) What is the velocity of the runner at this point?

Short answer

Answer: After 59.7 seconds, the runner has covered a distance of 360.3 meters, and her velocity is 6.3 m/s.

Step-by-step solution

a) Finding the distance covered after 59.7 seconds

First, we need to find the time taken for the runner to reach the final velocity of 6.3 m/s. As she starts from rest (initial velocity = 0) and accelerates at 1.25 m/s², we can use the equation: v = u + at Where: v is the final velocity (6.3 m/s) u is the initial velocity (0 m/s) a is the acceleration (1.25 m/s²) t is the time she takes to reach the final velocity. 6.3 = 0 + 1.25t t = (6.3 - 0) / 1.25 t = 5.04 s Now we find the distance covered during the acceleration phase: s_1 = ut + (1/2)at^2 s_1 = 0 * 5.04 + (1/2) * 1.25 * (5.04)^2 s_1 = 15.94 m Since the time given in the question is 59.7s, we can find the time spent during the constant velocity phase: t_2 = 59.7 - 5.04 t_2 = 54.66 s Now we find the distance covered during the constant velocity phase: s_2 = vt_2 s_2 = 6.3 * 54.66 s_2 = 344.36 m Finally, we find the total distance covered in 59.7 seconds: s_total = s_1 + s_2 s_total = 15.94 + 344.36 s_total = 360.3 m So after 59.7 seconds, the runner has covered 360.3 meters.

b) Finding the velocity of the runner at 59.7 seconds

As the runner has already reached her constant velocity of 6.3 m/s and continues running with the same velocity, her velocity at 59.7 seconds will also be: v = 6.3 m/s Hence, the velocity of the runner at 59.7 seconds is 6.3 m/s.

Key concepts

Constant Acceleration

Constant acceleration is a staple of kinematics in physics, representing a type of motion where the velocity of an object changes at a steady rate. This concept is central to understanding the movement of objects in a straight line when the force applied is uniform, and the resistance to motion (like friction or air resistance) is negligible or constant.

For example, our runner, who starts from rest, is experiencing a constant acceleration. This means her velocity increases by the same amount each second. During the acceleration phase, as there is a consistent change in speed, we can calculate other kinematic variables using this acceleration value, such as the time taken to reach a certain velocity and the distance covered during this acceleration.

In our scenario, the acceleration is given as \(1.25 \text{ m/s}^2\), indicating that the runner's speed increases by \(1.25 \text{ m/s}\) every second.

Velocity-Time Equation

The velocity-time equation is fundamental in connecting velocity, acceleration, and time within kinematic problems. When an object is undergoing constant acceleration, this equation is depicted as \(v = u + at\). This equation clarifies how the object's velocity (v) at any given time depends on its initial velocity (u), the acceleration (a), and the time (t) that has passed.

In our runner's case, she accelerates from an initial velocity (u) of 0 m/s, signifying a start from rest, to a final velocity (v) of \(6.3 \text{ m/s}\) using the constant acceleration provided. By rearranging this simple equation, we can solve for the time it takes for the runner to reach the final velocity during the acceleration phase of her motion. The knowledge of this time element is crucial as it serves as a bridge to calculating distances, which is another core concept in kinematics.

Distance Calculation

Distance calculation involves determining how far an object has traveled over a period, and it becomes slightly more intricate when the object undergoes different phases of motion, like our runner who first accelerates and then moves at a constant velocity. The distance covered during constant acceleration can be found using the formula \(s = ut + \frac{1}{2}at^2\), where 's' represents the distance, 'u' is the initial velocity, 'a' is the acceleration, and 't' is the time period of acceleration.

Once the runner reaches her top speed and continues with this constant velocity, the calculation simplifies to a multiplication of her constant speed by the elapsed time. Combining the distances from both the acceleration phase and the constant velocity phase gives the total distance covered. This stepwise approach helps students break down complex problems involving multiple phases of motion into manageable parts to get a clearer understanding of the whole scenario.