A Car in Constant Acceleration: Solving for Time Traveled

Imagine a scenario where a car starts from rest and accelerates at a constant rate. The goal is to determine the time it takes for the car to travel a certain distance given that the ratio of the distance traveled in the last second to the penultimate second is 4:3. This problem can be addressed through the equations of uniformly accelerated motion.

Understanding the Problem

Let's denote the total time of travel by (t), the initial velocity (u) by 0 (since the car starts from rest), and the constant acceleration by (a).

Formulas and Calculations

First, let's determine the distance traveled in the last second ((S_t)) and the penultimate second ((S_{t-1})).

Distance Traveled in the Last Second ((t))

The distance traveled in the last second can be given by:

[S_t frac{1}{2} a t^2 - frac{1}{2} a (t - 1)^2]

Expanding ((t - 1)^2):

[(t - 1)^2 t^2 - 2t 1]

Thus,

[S_t frac{1}{2} a t^2 - frac{1}{2} a (t^2 - 2t 1) frac{1}{2} a t^2 - frac{1}{2} a t^2 a t - frac{1}{2} a]

Simplifying, we get:

[S_t a t - frac{1}{2} a]

Distance Traveled in the Penultimate Second ((t - 1))

The distance traveled in the penultimate second is:

[S_{t-1} frac{1}{2} a (t - 1)^2 - frac{1}{2} a (t - 2)^2]

Expanding ((t - 1)^2) and ((t - 2)^2):

[(t - 1)^2 t^2 - 2t 1]

[(t - 2)^2 t^2 - 4t 4]

Thus,

[S_{t-1} frac{1}{2} a (t^2 - 2t 1) - frac{1}{2} a (t^2 - 4t 4) frac{1}{2} a t^2 - a t frac{1}{2} a - frac{1}{2} a t^2 2 a t - 2 a]

Simplifying, we get:

[S_{t-1} a t - frac{1}{2} a]

Setting Up the Ratio and Solving for Time

According to the problem, the ratio of the distances traveled in the last second and the penultimate second is 4:3:

[frac{S_t}{S_{t-1}} frac{4}{3}]

Substituting the expressions for (S_t) and (S_{t-1}):

[frac{a t - frac{1}{2} a}{a (t - 1) - frac{1}{2} a} frac{4}{3}]

Canceling (frac{1}{2} a) from the numerator and the denominator, assuming (a eq 0):

[frac{2 t - 1}{2 t - 3} frac{4}{3}]

Multiplying both sides by (2 t - 3):

[3(2 t - 1) 4(2 t - 3)]

Expanding and simplifying:

[6 t - 3 8 t - 12]

[-2 t -9]

[t frac{9}{2} 4.5 , text{seconds}]

Thus, the time traveled by the car is 4.5 seconds.

Summary

The solution to the problem involving a car starting from rest and moving with constant acceleration where the ratio of the distance traveled in the last second to the penultimate second is 4:3, shows that the time taken is 4.5 seconds. This example illustrates the application of equations of motion in solving real-world problems in physics.