High speed model cars are often run in circular paths by attaching them to a wire. Here we address a car attached to a steel wire of length \(8 \mathrm{~m}\). The car starts from rest and accelerates with a tangential acceleration \(a_{t}=0.5 \mathrm{~m} / \mathrm{s}^{2}\). (a) Find the speed of the car as a function of time. (b) Find the radial acceleration of the car as function of time. (c) At what speed is the radial acceleration 100 times larger than the tangential acceleration?

Tangential acceleration, denoted as \(\text{a}_{t}\), is the rate at which the tangential velocity of an object changes with time. In this problem, the tangential acceleration is given as \(\text{a}_{t} = 0.5 \text{ m/s}^2\). Think of tangential acceleration as how fast the car speeds up as it moves along the circular path. If you imagine the car speeding up at a constant rate, that's exactly what tangential acceleration measures.
In real-world terms, tangential acceleration affects how quickly the car goes from a standstill to its final speed. In this case, starting from rest with \(\text{v}_{0} = 0 \text{ m/s}\), we use the formula \(\text{v} = \text{v}_{0} + \text{a}_{t} t\).
This simple equation shows that the speed increases over time directly proportional to the tangential acceleration. For example, after 1 second at a tangential acceleration of 0.5 \text{ m/s}^2, the car's speed would be 0.5 \text{ m/s}. After 2 seconds, it would be 1 \text{ m/s}, and so on.

Kinematic equations relate the motion of objects under constant acceleration. These equations are fundamental in understanding how the position, velocity, and acceleration of an object change over time.
In the given problem, we start from the equation \(\text{v} = \text{v}_0 + \text{a}_{t} t\). Given the car starts from rest, \(\text{v}_0 = 0 \text{ m/s}\), and with a known tangential acceleration \(\text{a}_{t} = 0.5 \text{ m/s}^2\), we find: \[v(t) = 0 + 0.5 t = 0.5 t. \]
This equation tells us the speed of the car increases linearly with time. Kinematic equations are very reliable tools to predict motion in physics. They help us simplify complex motions and solve real-world problems effectively.

Radial acceleration, often called centripetal acceleration, is the acceleration that points towards the center of a circular path. It keeps the car moving in a curved trajectory. The formula for radial acceleration is \(\text{a}_r = \frac{\text{v}^2}{\text{r}}\).
For our car moving in a circle of radius \(\text{r} = 8 \text{ m}\), using the speed equation from Step 1, we calculate: \[ \text{a}_r(t) = \frac{(0.5 t)^2}{8} = \frac{0.25 t^2}{8} = \frac{t^2}{32} \text{ m/s}^2. \]
This shows how radial acceleration grows with the square of time. As the car speeds up, the force needed to keep it moving in a circle increases. Radial acceleration is crucial for understanding the forces acting on any object in circular motion.

Centripetal force is the force that causes an object to follow a curved path. It is directed towards the center of the curvature. Without this force, an object would continue in a straight line. For a car moving in a circle, the centripetal force comes from the tension in the steel wire.
We know from physics that centripetal force \(\text{F}_c\) is given by: \[ \text{F}_c = \text{m} \times \text{a}_r \] Our car's radial acceleration is \(\text{a}_r\) we've found earlier, and let's say the car has mass \(\text{m}\). Together, we can find the required centripetal force. For a given radial acceleration being 100 times the tangential acceleration: \[ \frac{\text{v}^2}{8} = 100 \times 0.5 \] Hence, \[ \text{v}^2 = 400. \text{So}, \text{v} = 20 \text{ m/s}. \]
At this speed, the radial force needed to keep the car in a circular path becomes significant. This demonstrates how speed impacts the forces at play in circular motion.