Problem 84 - Calculus and its applications

Question

A cheetah decelerates at a constant rate from \(50 \mathrm{~km} / \mathrm{hr}\) to a complete stop in \(20 \mathrm{sec}\). a) How fast is the cheetah moving after \(10 \mathrm{sec} ?\) b) How far has the cheetah traveled after 20 sec?

Step-by-Step Solution

Answer

After 10 seconds, the cheetah is moving at \(6.945 \text{ m/s}\). After 20 seconds, it has traveled 138.9 meters.

Step 1: Convert units for initial speed

First, convert the initial speed of the cheetah from \(50 \text{ km/hr}\) to meters per second (m/s).\[50 \text{ km/hr} = \frac{50 \times 1000}{3600} \text{ m/s} = \frac{50000}{3600} \text{ m/s} \approx 13.89 \text{ m/s}\]

Step 2: Determine the acceleration

Since the cheetah decelerates to a stop, use the formula for acceleration: \(a = \frac{v_f - v_i}{t}\), where \(v_f = 0\text{ m/s}\), \(v_i = 13.89\text{ m/s}\), and \(t = 20\text{ sec}\).\[a = \frac{0 - 13.89}{20} = -0.6945 \text{ m/s}^2\]

Step 3: Calculate velocity after 10 seconds

Use the formula \(v = v_i + at\) to find the velocity after 10 seconds, where \(t = 10\text{ sec}\), \(v_i = 13.89\text{ m/s}\), and \(a = -0.6945 \text{ m/s}^2\).\[v = 13.89 + (-0.6945) \times 10 = 13.89 - 6.945 = 6.945 \text{ m/s}\]

Step 4: Calculate the distance after 20 seconds

Use the equation for distance: \(s = v_i t + \frac{1}{2} a t^2\) for \(t = 20\text{ sec}\).\[s = 13.89 \times 20 + \frac{1}{2} \times (-0.6945) \times (20)^2\]Calculate:\[s = 277.8 - \frac{1}{2} \times 0.6945 \times 400\]\[s = 277.8 - 138.9 = 138.9 \text{ meters}\]

Key Concepts

Deceleration

Deceleration is the process of reducing speed over time, which is also known as negative acceleration. It occurs when a moving object slows down until it comes to a halt. In the exercise, the cheetah decelerates from 50 km/hr to a complete stop. To calculate deceleration, we use the formula for acceleration: - \(a = \frac{v_f - v_i}{t}\), where: - \(a\) is the acceleration (or deceleration in this context). - \(v_f\) is the final velocity, which is 0 in this case since the cheetah stops completely. - \(v_i\) is the initial velocity of 13.89 m/s (after conversion from km/hr). - \(t\) is the time taken to decelerate, which is 20 seconds here. Since the acceleration value is negative (-0.6945 m/s²), it indicates deceleration. Remember, the units and converting values are essential to ensure accuracy in these calculations.

Velocity Calculation

Velocity is how fast something is moving in a specific direction. Unlike simple speed, velocity has direction, making it a vector quantity. In the given problem, we need to calculate the velocity of the cheetah after 10 seconds of deceleration. We utilize the equation:- \(v = v_i + at\) - \(v\) is the velocity we are trying to find after 10 seconds. - \(v_i\) is the initial velocity of 13.89 m/s. - \(a\) is the deceleration, which is \(-0.6945\;m/s^2\). - \(t\) is the time, which in this part is 10 seconds.Plugging these values in, we get the velocity as 6.945 m/s. This means after 10 seconds, the cheetah has slowed down to this speed from its initial speed. The decrease in speed shows the impact of constant deceleration on the cheetah's motion.

Distance Calculation

Distance calculation helps us determine how far the cheetah travels over a certain period while it decelerates. In the problem, we want to find out the distance the cheetah covers in 20 seconds. We use the formula for distance covered with constant acceleration:- \(s = v_i t + \frac{1}{2} a t^2\) - \(s\) is the total distance traveled. - \(v_i\) is the initial velocity, which is once again 13.89 m/s. - \(t\) is the time, 20 seconds here. - \(a\) is the deceleration, \(-0.6945\;m/s^2\).Breaking it down:- First, calculate \(v_i t\): \(13.89 \times 20 = 277.8\;m\).- Then calculate \(\frac{1}{2} a t^2\): \(-\frac{1}{2} \times 0.6945 \times 400 = -138.9\;m\).Adding these, we find \(s = 277.8 - 138.9 = 138.9\;m\). Hence, the cheetah travels 138.9 meters while decelerating.