Question

We are told the particle starts between 10 m and 12 m at t = 0, passes 22 m in the 4th second, and reaches 55–60 m at t = 12 seconds. We need to find when it reaches 88 meters. Since motion is uniform, we'll calculate velocity = Δx / Δt. From t = 0 to t = 12 s, displacement = approx 46.5 m → velocity ≈ 3.875 m/s. Initial pos ≈ 11 m, final pos = 88 m ⇒ Δx = 77 m. Time = 77 / 3.875 ≈ 19.87 s. So, the particle reaches x = 88 m at approximately 19.9 seconds.

Answer 1

19.9 seconds

Answer 2

The problem describes the motion of a particle with constant velocity (uniform motion). We are given ranges for the initial position, a position at an intermediate time, and a position at a later time. We need to find the time when the particle reaches a specific position.

Let the position of the particle be given by $x(t) = x_0 + vt$, where $x_0$ is the initial position at $t=0$ and $v$ is the constant velocity.

Given information:

1. Initial position at $t=0$: $x_0 \in [10, 12]$ m.
2. Passes 22 m in the 4th second: This means $x(t_1) = 22$ m for some $t_1 \in (3, 4]$ s.
3. Reaches 55–60 m at $t=12$ seconds: $x(12) \in [55, 60]$ m.
4. Find $t$ when $x(t) = 88$ m.

To find a single approximate answer, we can use the average values of the given ranges: Average initial position: $x_0 = \frac{10 + 12}{2} = 11$ m. Average position at $t=12$ s: $x(12) = \frac{55 + 60}{2} = 57.5$ m.

Now, we can calculate the average velocity using these average values: $x(12) = x_0 + v \times 12$ $57.5 = 11 + 12v$ $12v = 57.5 - 11$ $12v = 46.5$ $v = \frac{46.5}{12} = 3.875$ m/s.

Let's check if this average velocity is consistent with the "passes 22 m in the 4th second" condition. Using $x_0 = 11$ m and $v = 3.875$ m/s: $x(t_1) = 11 + 3.875 t_1 = 22$ $3.875 t_1 = 11$ $t_1 = \frac{11}{3.875} \approx 2.8387$ s. This value of $t_1$ is not in the 4th second (which is $t \in (3, 4]$ s). This indicates that the given conditions might not be perfectly consistent for a single exact uniform motion, or we need to consider the full ranges.

However, if the question expects a single numerical answer, the approach of using average values is common. Let's proceed with the calculated average velocity $v = 3.875$ m/s and average initial position $x_0 = 11$ m.

We need to find the time $t$ when the particle reaches $x(t) = 88$ m. $x(t) = x_0 + vt$ $88 = 11 + 3.875t$ $3.875t = 88 - 11$ $3.875t = 77$ $t = \frac{77}{3.875}$ $t \approx 19.87096$ s

Rounding to one decimal place, $t \approx 19.9$ s.

Explanation of the solution:

1. Assume uniform motion $x(t) = x_0 + vt$.
2. Calculate average initial position $x_0 = (10+12)/2 = 11$ m.
3. Calculate average position at $t=12$ s as $x(12) = (55+60)/2 = 57.5$ m.
4. Determine average velocity $v = (x(12) - x_0) / 12 = (57.5 - 11) / 12 = 46.5 / 12 = 3.875$ m/s.
5. Set up the equation for reaching 88 m: $88 = x_0 + vt = 11 + 3.875t$.
6. Solve for $t$: $t = (88 - 11) / 3.875 = 77 / 3.875 \approx 19.9$ s.