A speedboat moving at 30.0 m / s approaches a no-wake buoy marker 100 m ahead. The pilot slows t

A speedboat moving at 30.0 m / s approaches a no-wake buoy...

Key Concepts

Uniform Acceleration

Uniform acceleration refers to motion where an object's acceleration remains constant over time. This concept simplifies the analysis of motion because it allows the use of specific kinematic equations that relate displacement, velocity, acceleration, and time directly. In problems like these, the constant deceleration enables a straightforward application of these equations, ensuring that the relationships between the various physical quantities are linear or quadratic in form.

Kinematic Equations

Kinematic equations are a set of formulas that connect the key variables—displacement, initial velocity, final velocity, acceleration, and time—in scenarios involving constant acceleration. These equations provide a systematic way to calculate unknown parameters when the other variables are known. They are essential for solving motion problems in physics, including those involving deceleration or acceleration over a fixed distance.

Quadratic Equation in Kinematics

When solving for time in kinematics, particularly when displacement is defined by a quadratic term in time (due to the acceleration term being multiplied by one-half), the resulting equation often takes the form of a quadratic equation. Solving this quadratic equation using methods such as the quadratic formula allows us to determine the time at which a moving object reaches a specific displacement.

Sign Conventions

Sign conventions are critical in kinematics for correctly interpreting the direction of motion and acceleration. When dealing with deceleration, the acceleration is given a negative value relative to the chosen positive direction of motion. This consistent approach to assigning signs ensures that the computations yield physically meaningful results, such as correctly predicting the reduction in speed over time.

Transcript

00:01 Here we have a speedboat moving at 30 meters per second towards a no wake buoy marker 100 meters ahead.

00:08 The pilot is slowing the boat down by applying and negative 3.5 million per second squared acceleration.

00:15 And we would like to know how long it takes the boat to reach the buoy and what the velocity of the boat is when it reaches the buoy.

00:23 So we have these ear equals 30 meters per second.

00:27 Delta x is 100 meters and the acceleration is negative 3.5 meters per second squared.

00:34 We would like to know the amount of time, so we would be working with an equation like the final equals the initial plus acceleration time still to time.

00:44 But that would probably be more useful once we have the delta t and re asked and we calculate the vf which were also asked for.

00:53 First, we must calculate the delta t value.

00:57 Well, we can consider the other equations we have you are equating to have her disposal are delta x equals v zero delta t plus 1/2 a delta t square.

01:13 So here we would know the value of a now you have delta x and the value of the zero and we will be solving.

01:24 For delta t, this would be a quadratic equation in terms of delta t, but it should still be solvable.

01:32 So let's try this method here.

01:35 Delta x we have 100 v zero, a 30 delta t plus 1/2 negative 3.5 delta t squared parents and then let's see what we have here.

01:51 We can rewrite this as delta t squared times.

01:57 What's 1/2 of negative? 3.5? that's going to be, um, well, negative seven divided by four, right, because 3.5 7/2 plus 30 delta t minus 100 equals zero.

02:15 So i'm just rewriting this in a form that won't be easier to start with the quadratic equation.

02:20 Sorry.

02:20 With the quadratic formula, the quadratic formula tells is that delta t is going to be equal to negative b native 30 closer, minus square root of b squared.

02:32 It's 900 minus four times a time.

02:40 See, don't forget that native sign there.

02:45 Then we divide by two in the women in the learning of sign for the a in great so luckily this four and this four cancel out.

02:56 So we got three positives on a three negative signs.

02:59 So we have one left behind there, so we'll have 900 minus 700 under the square root look like.

03:07 So we have a negative 30 closer minus the square of 700.

03:13 Don't wait.

03:14 Sorry.

03:14 900,000,700 which is 200 of course, divided by that would be a negative seven halves.

03:25 So i may be right this, but i bring the negative to top and bring the two to the top at 60 plus or minus.

03:35 And then we're going to have that two times the square root of 200 divided by seven...

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