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Chapter 2: Problem 33
A car moving with constant acceleration covered the distance between twopoints \(60.0 \mathrm{~m}\) apart in \(6.00 \mathrm{~s}\). Its speed as it passesthe second point was \(15.0 \mathrm{~m} / \mathrm{s} .(\mathrm{a})\mathrm{What}\) was the speed at the first point? (b) What was theacceleration? (c) At what prior distance from the first point was the car atrest? (d) Graph \(x\) vs. \(t\) and \(v\) vs. \(t\) for the car from rest\(\left(t_{1}=0 \mathrm{~s}\right)\).
Short Answer
Expert verified
Initial speed at the first point was 5.0 m/s. Acceleration was 1.67 m/s^2. Car was at rest 7.5 m before the first point.
Step by step solution
01
Define known variables
Given that the car covers a distance of 60.0 m with constant acceleration in 6.00 s and the speed at the second point is 15.0 m/s.
02
Use the equation of motion to find initial speed (V1)
The equation to use is: \[ s = \frac{ (V_1 + V_2) }{ 2 } * t \] where s = 60.0 m, V2 = 15.0 m/s, and t = 6.00 s. Rearrange the formula to solve for V1:\[ V_1 = \frac{(2s)}{t} - V_2 = \frac{(2 \times 60.0)}{6.00} - 15.0 = 5.0 \text{ m/s} \]
03
Use another motion equation to find acceleration (a)
Use the equation: \[ V_2 = V_1 + at \] Rearrange to solve for acceleration a:\[ a = \frac{(V_2 - V_1)}{t} = \frac{(15.0 - 5.0)}{6.00} = 1.67 \text{ m/s}^2 \]
04
Calculate prior distance where car was at rest
Use the equation: \[ V_2^2 = V_1^2 + 2as \]Substitute V2 = 5.0 m/s (initial speed), V_1 = 0 (rest), and a = 1.67 m/s^2. Solve for s (distance when the car was at rest):\[0 = V_1^2 + 2as_stopping \]Rearrange to solve for s:\[s = \frac{V_1^2}{2a} = \frac{5.0^2}{2 \times 1.67} = 7.5 \text{ m}\]
05
Graph the position vs. time and velocity vs. time
A position vs. time graph (x vs. t) for the car involves plotting the parabola: \[ x = V_1 t + \frac{1}{2}at^2 \]A velocity vs. time graph (v vs. t) is a linear graph: \[ v = V_1 + at \]
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
equations of motion
The equations of motion describe how a physical object's position, speed, and acceleration change over time. They are especially useful for problems involving constant acceleration. The three classic equations are:
1. \( v = u + at \): This calculates velocity (v) at time t given the initial velocity (u) and acceleration (a).
2. \( s= ut + \frac{1}{2}at^2 \): This calculates distance (s) traveled.
3. \( v^2 = u^2 + 2as \): This relates the final velocity to initial velocity, acceleration, and distance traveled.
In our exercise, these equations help determine initial speed, acceleration, and the location where the car was initially at rest.
initial velocity
To find the car's initial velocity (\( V_1 \)), we start from a given distance and final velocity after a specific time. The car covers 60.0 meters in 6.00 seconds, reaching 15.0 m/s at the end. Using the equation:
\[ s = \frac{(V_1 + V_2)}{2} \times t \]
where s = 60.0 m, V_2 = 15.0 m/s, and t = 6.00 s. Rearranging to solve for V_1:
\[ V_1 = \frac{2s}{t} - V_2 \rightarrow \frac{2 \times 60.0}{6.00} - 15.0 = 5.0 \text{ m/s} \]
Thus, the initial speed of the car was 5.0 m/s.
acceleration calculation
Acceleration (\( a \)) indicates how quickly the car's speed changes. Knowing the initial and final velocities and the time it took, we use:
\[ V_2 = V_1 + at \]
Rearranging to solve for acceleration:
\[ a = \frac{(V_2 - V_1)}{t} \rightarrow \frac{(15.0 - 5.0)}{6.00} = 1.67 \text{ m/s}^2 \]
So, the car's acceleration is 1.67 m/s^2.
distance calculation
We need to find what distance the car covered from a point of rest to reach its initial speed of 5.0 m/s. Using the equation:
\[ V_2^2 = V_1^2 + 2as \]
Here, V_2 is the initial speed in the previous calculation (5.0 m/s), and V_1 is 0 since the car was at rest. Rearranging to solve for s:
\[ s = \frac{V_1^2}{2a} \rightarrow \frac{5.0^2}{2 \times 1.67} = 7.5 \text{ m} \]
The car was at rest 7.5 meters before the calculated initial point.
graphical representation of motion
Graphs provide a clear visual representation of the car's motion. For position vs. time (x vs. t), use:
\[ x = V_1 t + \frac{1}{2}at^2 \]
which forms a parabola. For velocity vs. time (v vs. t), use:
\[ v = V_1 + at \]
which creates a linear graph. Plot these to observe acceleration's effects:
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