https://r2.erweima.ai/mathbot/image/17264460493478n10glf0.jpg

Let's break down and solve the problem given in the image:

The problem describes the motion of a particle along the x-axis with constant acceleration.

### Given:
- The position at \( t = 0 \) is \( x_0 = 30 \, \text{m} \).
- The initial velocity is \( v_0 = -15 \, \text{m/s} \) (since it moves in the negative x-direction).
- The velocity is decreasing at a rate of \( 1.5 \, \text{m/s}^2 \), which means the acceleration is \( a = 1.5 \, \text{m/s}^2 \) (positive since it opposes the motion and slows down the particle).

### Now, let's address each part of the problem:

#### a) La aceleración (The acceleration):
From the problem, the rate at which the velocity decreases is \( 1.5 \, \text{m/s}^2 \). Therefore, the acceleration is:
\[ a = +1.5 \, \text{m/s}^2 \]

#### b) El gráfico de velocidad contra tiempo (Velocity vs. time graph):
The velocity equation under constant acceleration is:
\[
v(t) = v_0 + a t
\]
Substituting the values:
\[
v(t) = -15 + 1.5t
\]
This is a linear equation, where the slope represents the acceleration, and the initial value at \( t = 0 \) is \( -15 \, \text{m/s} \). The graph would be a straight line with a positive slope of \( 1.5 \, \text{m/s}^2 \) and an intercept at \( -15 \, \text{m/s} \).

#### c) El gráfico de posición contra tiempo (Position vs. time graph):
The position function for uniformly accelerated motion is given by:
\[
x(t) = x_0 + v_0 t + \frac{1}{2} a t^2
\]
Substituting the known values:
\[
x(t) = 30 + (-15) t + \frac{1}{2} (1.5) t^2
\]
\[
x(t) = 30 - 15 t + 0.75 t^2
\]
This is a quadratic equation, meaning the position vs. time graph will be a parabola.

#### d) El tiempo que tarda la partícula en recorrer los primeros 75 m (Time to travel the first 75 m):
We are asked to find the time it takes for the particle to travel 75 meters. We need to solve for \( t \) when the displacement is 75 m.

The displacement equation is:
\[
\Delta x = x(t) - x_0
\]
We know \( \Delta x = 75 \, \text{m} \), so:
\[
75 = (-15) t + 0.75 t^2
\]
This simplifies to:
\[
0.75 t^2 - 15 t - 75 = 0
\]
Now, let's solve this quadratic equation to find the time \( t \). I'll compute the solution.The solutions to the quadratic equation are approximately:

\[
t = -4.14 \, \text{seconds} \quad \text{(not physically meaningful as time cannot be negative)}
\]
\[
t = 24.14 \, \text{seconds}
\]

Thus, the time it takes for the particle to travel the first 75 meters is approximately \( 24.14 \, \text{seconds} \).

---

Would you like more details on any part of the solution? Here are some follow-up questions to deepen understanding:

1. How is the sign of acceleration related to the direction of motion?
2. How would the graphs change if the initial velocity were positive?
3. What would happen if the acceleration were negative?
4. How do you interpret the intercepts in the velocity-time graph?
5. Can you calculate the velocity of the particle after 10 seconds?

**Tip:** In uniformly accelerated motion, the velocity-time graph is always a straight line, and its slope is the acceleration.

Un móvil se desplaza a lo largo del eje X con una aceleración constante. Si su posición para t = 0 es 30 m y se mueve en dirección X negativa con una rapidez de 15 m/s que está disminuyendo a razón de 1.5 m/s² cada segundo, determinar: a) La aceleración. b) El gráfico velocidad contra tiempo. c) El gráfico posición contra tiempo. d) El tiempo que tarda la partícula en recorrer los primeros 75 m.

Learn how to solve a kinematics problem involving constant acceleration, where the initial velocity is negative and decreases by 1.5 m/s². This problem covers the velocity vs. time and position vs. time graphs, and calculates the time to travel 75 m.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation