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Chapter 12: Problem 9

The speedometer on your car reads a steady 35 mph. Could you be accelerating? Explain.

Short Answer

Expert verified
Yes, if the car changes direction while maintaining 35 mph, it is accelerating.

Step by step solution

01

Understand Constant Speed

A steady speed of 35 mph means the speed of the car is not changing over time. This is indicated by the speedometer maintaining the same reading continuously.
02

Define Acceleration

Acceleration refers to the rate of change of velocity. An object is accelerating if there is a change in the speed or direction of its motion.
03

Consider Velocity and Direction

Even if the speed is constant, acceleration can still occur if there is a change in direction. In physics, velocity is a vector quantity, which means it involves both speed and direction. Therefore, turning results in a change in velocity.
04

Conclusion on Acceleration Possibility

With a steady speed of 35 mph, you could be accelerating if the direction of the car changes, as acceleration includes both changes in speed and direction. Therefore, even with a constant speed, a change in direction would indicate acceleration.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Speed
When we talk about a vehicle maintaining a constant speed, we're saying that the speedometer shows the same value over time. In this case, the car is moving at precisely 35 mph at every moment. This means the speed doesn't increase or decrease.
For a better understanding:
  • Constant speed means no acceleration due to changes in speed.
  • The velocity is steady because only speed, not direction, remains unchanged.
While it might seem straightforward, constant speed doesn't mean you can't be accelerating, as direction plays a significant role in defining acceleration.
Velocity
Velocity is an important concept in physics, and it's slightly more complex than just speed. Think of velocity as speed with an extra dimension: direction.
When defining velocity:
  • It's a vector quantity, meaning it includes both magnitude (how fast) and direction (where to).
  • A car traveling at 35 mph northbound has a different velocity than one traveling at 35 mph eastbound.
Velocity can change if either the speed or the direction changes, influencing whether an object is accelerating.
Change in Direction
Even with a constant speed, a car can experience acceleration if there is a change in direction. When you turn your car around a corner, for example, the direction of the car's movement changes.
Here's how it happens:
  • The speedometer might still read 35 mph, but the path you're taking is curving or shifting.
  • Since velocity depends on both speed and direction, any alteration in direction while maintaining speed leads to a change in velocity.
This shows acceleration due to a change in the direction of travel, even without speeding up or slowing down.
Vector Quantity
Understanding vectors is crucial to grasping concepts like velocity and acceleration.
  • Vectors are quantities that have both a magnitude and a direction.
  • Velocity is a vector because it tells you not just how fast something is going, but also in which direction.
  • This differs from speed, which is a scalar quantity 鈥� it only deals with magnitude.
In physics, changes in vectors can signify changes in acceleration, underscoring why both speed and direction matter for movement.

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Most popular questions from this chapter

Write a in the form \(\mathbf{a}=a_{\mathrm{T}} \mathbf{T}+a_{\mathrm{N}} \mathbf{N}\) without finding T and N. $$\mathbf{r}(t)=(1+3 t) \mathbf{i}+(t-2) \mathbf{j}-3 t \mathbf{k}$$

Solve the initial value problems for \(\mathbf{r}\) as a vector function of \(t.\) Differential equation:\(\frac{d \mathbf{r}}{d t}=\left(t^{3}+4 t\right) \mathbf{i}+t \mathbf{j}+2 t^{2} \mathbf{k}\)Initial condition: \(\quad \mathbf{r}(0)=\mathbf{i}+\mathbf{j}\)

a. Show that \(d \mathbf{B} / d s\) is orthogonal to \(\mathbf{T}\) and to \(\mathbf{B}\). b. Deduce from (a) that \(d \mathbf{B} / d s\) is parallel to \(\mathbf{N},\) so \(d \mathbf{B} / d s\) is a scalar multiple of N. Traditionally, we write $$\frac{d \mathbf{B}}{d s}=-\tau \mathbf{N}$$ The scalar \(\tau\) is called the torsion along the curve and may be positive, negative, or zero.

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves. $$\mathbf{r}(t)=(3 \sin t) \mathbf{i}+(3 \cos t) \mathbf{j}+4 t \mathbf{k}$$

The eccentricity of Earth's orbit is \(e=0.0167,\) so the orbit is nearly circular, with radius approximately \(150 \times 10^{6} \mathrm{km}\). Find, in units of \(\mathrm{km}^{2} /\) sec, the rate \(d A /\) d \(t\) satisfying Kepler's second law.
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