Problem Statement
Pathfinder KinematicsThis problem utilizes basic one-dimensional kinematic equations. We analyze the motion by breaking it into two distinct phases: uniform acceleration and uniform deceleration.
Step-by-Step Derivation
Phase 1: Acceleration
First, we need to look at the acceleration phase. The car starts from rest, meaning initial velocity $u = 0$. Let the time spent in this phase be $t_1$. The maximum velocity $v_{max}$ is reached precisely at the end of this acceleration phase.
$$v_{max} = 0 + \alpha t_1 \implies t_1 = \frac{v_{max}}{\alpha}$$
Phase 2: Deceleration
Next, we analyze the braking phase. The car begins this phase with velocity $v_{max}$ and comes to a complete halt, meaning final velocity is $0$. Let the time spent decelerating be $t_2$.
$$0 = v_{max} – \beta t_2 \implies t_2 = \frac{v_{max}}{\beta}$$
Finding Maximum Velocity
We are given that the total time elapsed is $t$. Therefore, $t = t_1 + t_2$. Substituting our values for time:
$$t = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta}$$
$$t = v_{max} \left( \frac{\alpha + \beta}{\alpha \beta} \right)$$
$$v_{max} = \frac{\alpha \beta}{\alpha + \beta} t$$
Notice how doing this algebraically takes a few steps? You can completely bypass the algebra by drawing a Velocity-Time (v-t) graph.
The motion forms a triangle with the time axis as the base ($t$) and the peak representing height ($v_{max}$). The time intervals on the base are simply the height divided by the respective slopes (accelerations):
$$t = \frac{v_{max}}{\alpha} + \frac{v_{max}}{\beta}$$
This visual method eliminates the need to write out the individual phase equations and gets you directly to the final relationship in under 10 seconds. This is where most students make a mistake by overcomplicating the math!
