**Physics Problem Example****Problem:**A truck accelerates at a constant acceleration of 5.2 m/s² from rest. How far does the truck travel in the first 3.9 seconds, in meters?**Solution:**To determine the distance traveled by the truck, we can use the kinematic equation for distance under constant acceleration:\[ d = v_i t + \frac{1}{2} a t^2 \]Where:- \( d \) = distance- \( v_i \) = initial velocity (0 m/s in this case, as the truck starts from rest)- \( t \) = time (3.9 seconds)- \( a \) = acceleration (5.2 m/s²)Since the truck starts from rest, the initial velocity \( v_i \) is 0. Thus, the equation simplifies to:\[ d = \frac{1}{2} a t^2 \]Plugging the given values into the equation:\[ d = \frac{1}{2} \times 5.2 \, \text{m/s}^2 \times (3.9 \, \text{s})^2 \]\[ d = 0.5 \times 5.2 \times 15.21 \]\[ d = 2.6 \times 15.21 \]\[ d \approx 39.55 \, \text{meters} \]**Conclusion:**The truck travels approximately 39.55 meters in the first 3.9 seconds.

Given: The acceleration of the truck is 5.2 m/s2. The itnial speed of the truck is 0 m/s. The time…

Answered: A truck accelerates at a constant…

A truck accelerates at a constant acceleration of 5.2 m/s² from rest. How far does the truck travel in the first 3.9 s, in meters?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Distance and Speed

The speed, distance and time relation forms the basis of study of motion. You will need these concepts in the study of Mechanics. This concept is applicable for students who are studying the following courses –

Question

**Physics Problem Example**

**Problem:**
A truck accelerates at a constant acceleration of 5.2 m/s² from rest. How far does the truck travel in the first 3.9 seconds, in meters?

**Solution:**
To determine the distance traveled by the truck, we can use the kinematic equation for distance under constant acceleration:

\[ d = v_i t + \frac{1}{2} a t^2 \]

Where:
- \( d \) = distance
- \( v_i \) = initial velocity (0 m/s in this case, as the truck starts from rest)
- \( t \) = time (3.9 seconds)
- \( a \) = acceleration (5.2 m/s²)

Since the truck starts from rest, the initial velocity \( v_i \) is 0. Thus, the equation simplifies to:

\[ d = \frac{1}{2} a t^2 \]

Plugging the given values into the equation:

\[ d = \frac{1}{2} \times 5.2 \, \text{m/s}^2 \times (3.9 \, \text{s})^2 \]

\[ d = 0.5 \times 5.2 \times 15.21 \]

\[ d = 2.6 \times 15.21 \]

\[ d \approx 39.55 \, \text{meters} \]

**Conclusion:**
The truck travels approximately 39.55 meters in the first 3.9 seconds.

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