A car stops 4 seconds after the application of the brakes while covering a rectilinear stretch 337 feet long. If the motion occurred with a constant acceleration a, determine the initial speed of the car and the acceleration a. Express v in mile-per-hour (mph) and a in terms of g, the acceleration of gravity. S

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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**Problem Statement:**

A car stops 4 seconds after the application of the brakes while covering a rectilinear stretch 337 feet long. If the motion occurred with a constant acceleration \( a_c \), determine the initial speed \( v_o \) of the car and the acceleration \( a_c \). Express \( v_o \) in mile-per-hour (mph) and \( a_c \) in terms of \( g \), the acceleration of gravity.

**Diagram Explanation:**

The image includes an illustration of a car moving on a straight path, labeled with a distance \( s \). An arrow indicates the direction of motion, suggesting that the car is braking to a stop over the specified distance of 337 feet. 

To solve the problem, we can use the kinematic equation:
\[ s = v_o t + \frac{1}{2} a_c t^2 \]

Where:
- \( s = 337 \) feet
- \( t = 4 \) seconds

First, solve for \( v_o \) and \( a_c \) using the given conditions.
Next, to convert \( v_o \) from feet per second to miles per hour, use the conversion factor:
\[ 1 \text{ ft/s} = 0.6818 \text{ mph} \]

Finally, express \( a_c \) in terms of \( g \) (where \( g \approx 32.2 \) ft/s²).
Transcribed Image Text:**Problem Statement:** A car stops 4 seconds after the application of the brakes while covering a rectilinear stretch 337 feet long. If the motion occurred with a constant acceleration \( a_c \), determine the initial speed \( v_o \) of the car and the acceleration \( a_c \). Express \( v_o \) in mile-per-hour (mph) and \( a_c \) in terms of \( g \), the acceleration of gravity. **Diagram Explanation:** The image includes an illustration of a car moving on a straight path, labeled with a distance \( s \). An arrow indicates the direction of motion, suggesting that the car is braking to a stop over the specified distance of 337 feet. To solve the problem, we can use the kinematic equation: \[ s = v_o t + \frac{1}{2} a_c t^2 \] Where: - \( s = 337 \) feet - \( t = 4 \) seconds First, solve for \( v_o \) and \( a_c \) using the given conditions. Next, to convert \( v_o \) from feet per second to miles per hour, use the conversion factor: \[ 1 \text{ ft/s} = 0.6818 \text{ mph} \] Finally, express \( a_c \) in terms of \( g \) (where \( g \approx 32.2 \) ft/s²).
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