Problem 13 - Applied Calculus

Question

A car initially going 50 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 5 seconds. (a) Graph the velocity from \(t=0\) to \(t=5\). (b) How far does the car travel? (c) How far does the car travel if its initial velocity is doubled, but it brakes at the same constant rate?

Step-by-Step Solution

Answer

(a) Graph is a line decreasing from 50 ft/sec to 0 ft/sec over 5 seconds. (b) The car travels 125 feet. (c) With doubled initial velocity, the car travels 500 feet.

Step 1: Determine Deceleration

The car comes to a stop in 5 seconds from an initial velocity of 50 ft/sec. Use the formula for constant acceleration (or deceleration) which is given by \( v = v_0 + at \). Since the final velocity \( v = 0 \) at \( t = 5 \) seconds, we can rearrange to solve for \( a \): \( 0 = 50 + a \times 5 \). Thus, \( a = -10 \) ft/sec².

Step 2: Graph the Velocity

The graph of velocity versus time is a straight line starting at 50 ft/sec at \( t=0 \) and ending at 0 ft/sec at \( t=5 \). The slope of the line is the constant deceleration from Step 1, \(-10\) ft/sec².

Step 3: Calculate Distance Travelled

To find the distance travelled, use the formula for distance in constant acceleration: \( d = v_0 t + \frac{1}{2} a t^2 \). Substitute the values: \( v_0 = 50 \) ft/sec, \( a = -10 \) ft/sec², \( t = 5 \) seconds. \( d = 50 \times 5 + \frac{1}{2} (-10) \times 5^2 = 250 - 125 = 125 \) feet.

Step 4: Analyze Doubling Initial Velocity

If the initial velocity is doubled to 100 ft/sec, use the same braking rate \(-10\) ft/sec². The time to stop becomes \( t = \frac{v_0}{|a|} = \frac{100}{10} = 10 \) seconds. Recalculate the distance using \( d = v_0 t + \frac{1}{2} a t^2 \): \( d = 100 \times 10 + \frac{1}{2}(-10) \times 10^2 = 1000 - 500 = 500 \) feet.

Key Concepts

Velocity-Time Graph

A velocity-time graph is a graphical representation that shows how the velocity of an object changes with time. This graph is crucial in understanding how an object accelerates or decelerates over a period. It is represented with time on the x-axis and velocity on the y-axis. In our scenario,

• The graph starts at an initial velocity of 50 ft/sec at time zero.
• It linearly decreases to a velocity of 0 ft/sec at 5 seconds.

The linear decrease indicates a constant deceleration. The slope of the line on this graph represents the rate of acceleration or deceleration. In this case, the slope is negative, representing a deceleration of (−10) ft/sec². This is a key aspect of understanding how the car slows to a stop over 5 seconds.

Deceleration

Deceleration is the rate at which an object slows down. It can be thought of as negative acceleration. In this exercise, the car experiences a constant deceleration as it comes to a complete stop.

• The formula used is the same as acceleration: \( v = v_0 + at \)
• With final velocity \( v = 0 \) and initial velocity \( v_0 = 50 \) ft/sec.
• The car takes 5 seconds to stop, allowing us to calculate the deceleration as \(a = -10\) ft/sec².

This deceleration means the car decreases its velocity by 10 ft/sec each second until it reaches zero. Understanding deceleration in terms of constant negative acceleration is important because it directly relates to how an object slows down under braking.

Distance Traveled

Distance traveled during constant acceleration can be calculated using the formula: \(d = v_0 t + \frac{1}{2} a t^2\). This equation helps determine how far an object moves under constant acceleration or deceleration. In this exercise,

• The initial velocity \(v_0\) was 50 ft/sec.
• Time \(t\) is 5 seconds.
• Deceleration \(a\) is −10 ft/sec².

Substituting these values, the distance the car traveled is calculated as 125 feet. When initial velocity is increased, as in part (c) of the exercise, it demonstrates how dramatically distance can increase even if the braking rate stays constant. Doubling initial velocity to 100 ft/sec results in the car traveling 500 feet before stopping.

Initial Velocity

Initial velocity is the speed of an object right at the start before any forces such as acceleration or deceleration affect it. In the exercise, the car begins with an initial velocity of 50 ft/sec. Initial velocity is a critical factor in determining other variables like stopping time and distance.

• It provides a starting point for calculating how the object will move over time.
• When the initial velocity is doubled to 100 ft/sec, the car takes longer to stop, at 10 seconds.
• This increased time also means a greater distance is covered, which is 500 feet.

Understanding initial velocity helps predict how changes to this parameter affect overall motion, enabling better planning and analysis in real-world situations.