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A cheetah can accelerate from rest to \(25.0 \mathrm{~m} / \mathrm{s}\) in \(6.00 \mathrm{~s}\). Assuming that the cheetah moves with constant acceleration, what distance does it cover in the first \(3.00 \mathrm{~s}\) ?

Short Answer

Expert verified
The cheetah covers 18.8 m in the first 3 seconds.

Step by step solution

01

Determine Acceleration

The cheetah accelerates from rest to a velocity of \(25.0 \; \mathrm{m/s}\) in \(6.00 \; \mathrm{s}\). Use the formula for constant acceleration \(v = u + at\) to solve for \(a\). Here, \(v = 25.0 \; \mathrm{m/s}\), \(u = 0\), and \(t = 6.00 \; \mathrm{s}\).\[a = \frac{v - u}{t} = \frac{25.0 - 0}{6.00} = 4.17 \; \mathrm{m/s^2}.\]
02

Calculate Distance in 3 Seconds

Now that we know the acceleration is \(4.17 \; \mathrm{m/s^2}\), use the formula for distance traveled given the initial velocity, acceleration, and time: \(s = ut + \frac{1}{2}at^2\). For the first \(3.00 \; \mathrm{s}\), \(u = 0\) and \(t = 3.00 \; \mathrm{s}\), so:\[s = 0 + \frac{1}{2} \times 4.17 \times (3.00)^2 = 18.8 \; \mathrm{m}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Acceleration
Constant acceleration is a fundamental concept in kinematics. It occurs when an object's velocity changes at a consistent rate over time. Imagine a car speeding up at a steady pace when you press the accelerator pedal all the way to the floor. This predictable increase in speed is a prime example of constant acceleration.

For the cheetah in our exercise, starting from rest and reaching a velocity of 25.0 meters per second in 6 seconds demonstrates constant acceleration. To determine this rate of change in velocity, we use the formula:
  • \( v = u + at \)
Where:
  • \( v \) is the final velocity (25.0 m/s)
  • \( u \) is the initial velocity (0 m/s since it starts from rest)
  • \( a \) is the acceleration
  • \( t \) is the time duration (6.00 s)
By rearranging the formula to solve for acceleration \( a \), we find that the cheetah has a constant acceleration of 4.17 m/s².
Distance Calculation
Calculating how far an object travels under constant acceleration involves another basic kinematic equation. This is useful when you need to know the distance covered within a certain time frame while an object is accelerating. In our scenario, we need to find the distance the cheetah covers in the first 3 seconds. Here’s the formula we use:
  • \( s = ut + \frac{1}{2}at^2 \)
In this equation:
  • \( s \) is the distance traveled
  • \( u \) is the initial velocity (0 m/s in this case since we started from rest)
  • \( a \) is the constant acceleration (4.17 m/s²)
  • \( t \) represents the time duration in question (3.00 seconds here)
Substituting these values into the formula, we calculate:
  • \( s = 0 + \frac{1}{2} \times 4.17 \times (3.00)^2 = 18.8 \, \mathrm{m} \)
Naturally, this means that the cheetah covers 18.8 meters in the first 3 seconds under constant acceleration.
Velocity
Velocity, another fundamental concept in kinematics, describes how fast an object is moving in a specific direction. It is a vector quantity, meaning it has both magnitude and direction. Unlike speed, which is scalar and only considers how fast an object is moving, velocity must specify the direction of movement.

In our cheetah exercise, the velocity at the end of 6 seconds is 25.0 m/s. This final velocity plays a crucial role in finding acceleration when the cheetah starts from rest. Understanding how velocity changes over time when an object is experiencing constant acceleration helps us predict motion paths and distances covered accurately.

By visualizing velocity as a straight-line increase in our graph of time versus velocity, we can see the nature of constant acceleration. As all velocity starts at 0 and steadily increases to 25.0 m/s over a period of 6 seconds, it's clear that the cheetah gains velocity in equal increments each second, thanks to its constant acceleration.

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Most popular questions from this chapter

At the starting gun a runner accelerates from rest at \(1.9 \mathrm{~m} / \mathrm{s}^{2}\) for \(2.2 \mathrm{~s}\). What is the runner's speed \(2.0 \mathrm{~s}\) after she starts running?

Can an object that is moving upward be in free fall?

Assume that the brakes in your car create a constant deceleration of \(4.2 \mathrm{~m} / \mathrm{s}^{2}\) regardless of how fast you are driving. (a) If you double your driving speed from \(16 \mathrm{~m} / \mathrm{s}\) to \(32 \mathrm{~m} / \mathrm{s}\), does the time required to come to a stop increase by a factor of 2 or a factor of 4 ? Explain. Verify your answer to part (a) by calculating the stopping times for initial speeds of (b) \(16 \mathrm{~m} / \mathrm{s}\) and (c) \(32 \mathrm{~m} / \mathrm{s}\).

The position-time equation for a certain train is $$ x_{\mathrm{f}}=2.1 \mathrm{~m}+(8.3 \mathrm{~m} / \mathrm{s}) t+\left(2.6 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2} $$ (a) What is the initial velocity of this train? (b) What is its acceleration?

Think \& Calculate A boat is cruising in a straight line at a constant speed of \(2.6 \mathrm{~m} / \mathrm{s}\) when it is shifted into neutral. After coasting \(12 \mathrm{~m}\), the boat is put back into gear and resumes cruising at the reduced constant speed of \(1.6 \mathrm{~m} / \mathrm{s}\). (a) Assuming that the acceleration was constant during coasting, how much time did it take for the boat to coast the \(12 \mathrm{~m}\) ? (b) What was the boat's acceleration while it was coasting? (c) When the boat had coasted for \(6.0 \mathrm{~m}\), was its speed \(2.1 \mathrm{~m} / \mathrm{s}\), more than \(2.1 \mathrm{~m} / \mathrm{s}\), or less than \(2.1 \mathrm{~m} / \mathrm{s}\) ? Explain.

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