Answer
Result See proof
Work Step by Step
Step 1 Before we answer the question, it is important to first understand how the acceleration vectors mathematically work. Here, it is written as the following equation: \[ \mathbf{a} = \frac{d^2s}{dt^2}\mathbf{T} + \kappa\left(\frac{dt}{ds}\right)^2\mathbf{N} \] (1) Let us also recall the equation above is rooted in the original equation, given below: \[ \mathbf{a} = \frac{d}{dt}\left(\frac{ds}{dt}\mathbf{T}\right) \] (2) Where our acceleration is divided into two components: the tangential one (which has the direction of \(\mathbf{T}\)), and the normal one (which has the direction of \(\mathbf{N}\)). As such, the acceleration that we feel in real life also denotes these components, where we can feel it pushing us within the curve. Step 2 Now, to answer this question, when a car tries to go down a hill or do a sharp turn, one feels a push to the opposite direction of the curve because you are reacting to the car itself. For example, when a car does a sharp turn to the left, the Normal Acceleration component is also going to the left. However, you'll actually feel yourself moving to the right since what you did is the reaction to the car driving to the left. The same is true when one feels lighter when moving down in the rollercoaster. While the acceleration due to gravity remains constant, the force holding you down (the coaster) is not reacting to you going down. Hence, you feel lighter or feel like flying. As such, the sudden change in velocity made by the car or the coaster makes you feel the opposite due to the relative speeds both you and these vehicles are feeling. These are the basic principles of inertia. Step 3 Moving forward, upon changing the vehicle's velocity, according to equation (1), there would be consequent effects on the components of acceleration, and thus, acceleration itself. Looking at equation (2), we see that upon doubling the speed, we would also double both \(\frac{d^2s}{dt^2}\) and consequently, \(\frac{ds}{dt}\). As such, when the vehicle's velocity is doubled, the tangential component of acceleration would feel double the acceleration, while the Normal component, since it is squared, would feel four times the acceleration. Consequently, when the vehicle's velocity is halved, it is also evident that the tangential component would also halve. In contrast, the Normal component would be reduced to one-fourth of its original value. As such, the forces would also adjust accordingly, since \(\mathbf{F} = m\mathbf{a}\).
