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.

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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.