Problem 1:   The world's fastest land animal, the cheetah, can accelerate at 9.05 m/s2 to a top speed of 30.5 m/s. A cheetah observes a passing gazelle traveling at 19.5 m/s and begins to chase it. 
Part (a) How long, in seconds, does it take the cheetah to reach its maximum velocity, assuming its acceleration is constant?
Numeric   : A numeric value is expected and not an expression.
t
t
 = __________________________________________s


Part (b) How far, in meters, has the cheetah traveled, dmax,vel, when it reaches it maximum velocity?
Numeric   : A numeric value is expected and not an expression.
dmax,vel = __________________________________________


Part (c) Assume that d
d
 is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d
d
, the cheetah's maximum velocity vmax
v
max
, and the velocity of the gazelle vg
v
g
 for the time, tc
t
c
, it takes the cheetah to catch the gazelle. 
Expression   :
tc
t
c
 = __________________________________________

Select from the variables below to write your expression. Note that all variables may not be required.
β, γ, θ, b, d, g, h, j, k, m, n, P, S, vcmax, vg

Part (d) What is the numeric value for this time, tc
t
c
 in seconds, assuming the cheetah is 31 meters away when it reaches its maximum velocity?
Numeric   : A numeric value is expected and not an expression.
tc
t
c
 = __________________________________________s

Let's go through each part of the problem step by step.

### Part (a) Time to reach maximum velocity

We are given:
- Acceleration of the cheetah: \( a = 9.05 \, \text{m/s}^2 \)
- Final velocity: \( v = 30.5 \, \text{m/s} \)

To find the time \( t \) it takes to reach maximum velocity, we use the equation of motion:
\[
v = a \cdot t
\]
Solving for \( t \):
\[
t = \frac{v}{a} = \frac{30.5 \, \text{m/s}}{9.05 \, \text{m/s}^2}
\]

Let's calculate \( t \).

\[
t \approx 3.37 \, \text{seconds}
\]

So, the cheetah takes **3.37 seconds** to reach its maximum velocity.

### Part (b) Distance traveled when reaching maximum velocity

To find the distance \( d \) traveled while the cheetah accelerates to its maximum speed, we use the following kinematic equation:
\[
d = \frac{1}{2} a t^2
\]
From part (a), we have \( t = 3.37 \, \text{seconds} \), and the acceleration is \( a = 9.05 \, \text{m/s}^2 \). Plugging these values in:

\[
d = \frac{1}{2} \cdot 9.05 \, \text{m/s}^2 \cdot (3.37)^2
\]

Let's calculate the distance \( d \).

\[
d \approx 51.44 \, \text{meters}
\]

So, the cheetah travels **51.44 meters** while reaching its maximum velocity.

### Part (c) Time for the cheetah to catch the gazelle

Now, the cheetah has reached its maximum speed \( v_{\text{max}} \), and the gazelle is moving at a constant speed \( v_g = 19.5 \, \text{m/s} \). The cheetah needs to cover the distance \( d \) between them.

The relative speed at which the cheetah is closing the gap is:
\[
v_{\text{relative}} = v_{\text{max}} - v_g
\]

The time \( t_c \) it takes for the cheetah to catch the gazelle is:
\[
t_c = \frac{d}{v_{\text{relative}}}
\]
So the expression for \( t_c \) is:
\[
t_c = \frac{d}{v_{\text{max}} - v_g}
\]

### Part (d) Numeric value for \( t_c \)

We are given:
- \( d = 31 \, \text{meters} \)
- \( v_{\text{max}} = 30.5 \, \text{m/s} \)
- \( v_g = 19.5 \, \text{m/s} \)

Using the formula from part (c):
\[
t_c = \frac{31 \, \text{meters}}{30.5 \, \text{m/s} - 19.5 \, \text{m/s}} = \frac{31}{11}
\]

Let's calculate \( t_c \).

\[
t_c \approx 2.82 \, \text{seconds}
\]

So, the time it takes for the cheetah to catch the gazelle is **2.82 seconds**.

---

Would you like more details on any of these steps?

### Related questions:
1. How would the distance change if the cheetah had a higher acceleration?
2. What happens if the gazelle starts accelerating as well?
3. How long would it take the cheetah to catch the gazelle if it were 50 meters away?
4. What would happen if the gazelle was faster than the cheetah’s maximum speed?
5. Can we calculate the average velocity of the cheetah during its acceleration phase?

### Tip:
When solving motion problems, carefully choose the kinematic equation that matches the known variables for a more straightforward solution.

The world's fastest land animal, the cheetah, can accelerate at 9.05 m/s² to a top speed of 30.5 m/s. A cheetah observes a passing gazelle traveling at 19.5 m/s and begins to chase it. Part (a) How long, in seconds, does it take the cheetah to reach its maximum velocity, assuming its acceleration is constant? Part (b) How far, in meters, has the cheetah traveled when it reaches its maximum velocity? Part (c) Derive an expression for the time it takes the cheetah to catch the gazelle. Part (d) What is the numeric value for this time, assuming the cheetah is 31 meters away when it reaches its maximum velocity?

Solve a kinematic problem involving the cheetah chasing a gazelle. Learn how to calculate the time it takes for the cheetah to reach its maximum speed, the distance covered, and the time to catch the gazelle. Includes step-by-step solutions using key formulas like v = a * t, and suitable for Grades 10-12.

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