Question

You are driving a car that has a maximum acceleration of a. The magnitude of the maximum deceleration is also a. What is the maximum distance that you can travel in time T, assuming that you begin and end at rest?

• A. 2aT²
• B. aT²
• C. aT²/4
• D. aT²/8

Answer 1

Answer: (C)

aT²/4

Answer 2

To find the maximum distance traveled, we need to maximize the area under the velocity-time (v-t) graph. Since the car starts and ends at rest, the v-t graph will start at $v=0$ and end at $v=0$. To maximize the area (distance), the car should accelerate at its maximum rate 'a' for some time, reach a maximum velocity ($v_{max}$), and then decelerate at its maximum rate 'a' to come to rest. This forms a triangular v-t graph.

Let $t_1$ be the time taken to accelerate from rest to $v_{max}$. Using the equation $v = u + at$: $v_{max} = 0 + at_1 \implies v_{max} = at_1$

Let $t_2$ be the time taken to decelerate from $v_{max}$ to rest. Using the equation $v = u + at$ (where 'a' is deceleration, so we use -a): $0 = v_{max} - at_2 \implies v_{max} = at_2$

From these two equations, we see that $at_1 = at_2$, which implies $t_1 = t_2$. The total time given is $T = t_1 + t_2$. Substituting $t_1 = t_2$, we get $T = t_1 + t_1 = 2t_1$. Therefore, $t_1 = T/2$. And consequently, $t_2 = T/2$.

Now, we can find the maximum velocity attained: $v_{max} = a t_1 = a(T/2)$.

The total distance traveled is the area of the triangular v-t graph: Distance $S = \frac{1}{2} \times \text{base} \times \text{height}$ $S = \frac{1}{2} \times T \times v_{max}$ Substitute the expression for $v_{max}$: $S = \frac{1}{2} \times T \times (aT/2)$ $S = \frac{aT^2}{4}$

The maximum distance that can be traveled in time T is $\frac{aT^2}{4}$.