What is the formula of a cheetah weights in hiding behind a bush the cheetah spot the glazel running pass at 10 m per second at the instant of glazel passes the cheetah the cheetah accelerates from a rest at 4 m and second square to catch the glazel how long does it take the cheetah to catch the glazel what is the displacement of the glazel and cheetah

This problem can be solved using equations of motion for both the cheetah and the gazelle.

### Given Information:
- **Speed of the gazelle**: \(v_g = 10 \, \text{m/s}\) (constant)
- **Cheetah's initial velocity**: \(u_c = 0 \, \text{m/s}\) (at rest)
- **Cheetah's acceleration**: \(a_c = 4 \, \text{m/s}^2\)
- **Cheetah's displacement**: \(d_c\)
- **Gazelle's displacement**: \(d_g\)
- **Time taken to catch the gazelle**: \(t\)

We need to find:
1. The time it takes for the cheetah to catch the gazelle.
2. The displacement of the gazelle and the cheetah when the cheetah catches the gazelle.

### Step 1: Set up the equations of motion

1. **Gazelle's motion**:
   Since the gazelle is moving with a constant velocity, its displacement after time \(t\) is given by:
   \[
   d_g = v_g \cdot t = 10t
   \]

2. **Cheetah's motion**:
   The cheetah starts from rest and accelerates with a constant acceleration. The displacement after time \(t\) for uniformly accelerated motion is given by:
   \[
   d_c = u_c t + \frac{1}{2} a_c t^2 = 0 + \frac{1}{2} \cdot 4 \cdot t^2 = 2t^2
   \]

### Step 2: Equate the displacements
Since the cheetah catches the gazelle, their displacements are equal at the point of catching:
\[
d_g = d_c
\]
\[
10t = 2t^2
\]

### Step 3: Solve for time \(t\)
Rearranging the equation:
\[
2t^2 - 10t = 0
\]
Factor the equation:
\[
t(2t - 10) = 0
\]
Thus, \(t = 0 \, \text{or} \, t = 5 \, \text{seconds}\).

Since \(t = 0\) is the initial time, we take \(t = 5 \, \text{seconds}\).

### Step 4: Find the displacement of the gazelle and cheetah
Substitute \(t = 5 \, \text{seconds}\) into the displacement equations:

- Gazelle's displacement:
  \[
  d_g = 10 \cdot 5 = 50 \, \text{meters}
  \]

- Cheetah's displacement:
  \[
  d_c = 2 \cdot (5)^2 = 2 \cdot 25 = 50 \, \text{meters}
  \]

### Final Answer:
1. The time it takes for the cheetah to catch the gazelle is **5 seconds**.
2. The displacement of both the gazelle and the cheetah is **50 meters**.

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Would you like more details or have any questions about this solution? Here are some related questions you might consider:

1. How would the problem change if the cheetah had a different acceleration?
2. What if the gazelle's speed were variable instead of constant?
3. How would air resistance affect the cheetah's ability to catch the gazelle?
4. Can the time to catch the gazelle be reduced by starting with an initial velocity?
5. How does this problem illustrate the concept of relative motion?

**Tip:** When solving problems involving acceleration, always consider the initial velocity and whether the motion is uniformly accelerated or not!

Explore how to solve a kinematics problem where a cheetah catches a gazelle. Given the gazelle's constant speed and the cheetah's acceleration, learn to calculate the time and displacement involved. Suitable for students in Grades 9-12.

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