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Chapter 2: Problem 17

The position of the front bumper of a test car under microprocessor control is given by \(x(t)=2.17 \mathrm{~m}+\left(4.80 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}-\) \(\left(0.100 \mathrm{~m} / \mathrm{s}^{6}\right) t^{6} .\) (a) Find its position and acceleration at the instants when the car has zero velocity. (b) Draw \(x-t, v_{x}-t,\) and \(a_{x}-t\) graphs for the motion of the bumper between \(t=0\) and \(t=2.00 \mathrm{~s}\)

Short Answer

Expert verified
The exercise requires various applications of calculus to derive the velocity and acceleration functions from the position function, then solve for the instances of zero velocity. With those times, determine position and acceleration at these times, and then plot the motion graphs for the given time range.

Step by step solution

01

Understand the given function

The given function, \(x(t)=2.17m+(4.80m/s^2)t^2-(0.100m/s^6)t^6\), represents the position of the front bumper of a test car at any given time 't'. It is in meters.
02

Derive the velocity function

Use the derivative to convert the position function into a velocity function. Differentiate the position function \(x(t)\). The derivative of \(x(t)\) will give \(v(t)\). Apply the power rule, which states that the derivative of \(x^n\), is \(n\*x^{n-1}\).
03

Solve v(t) for zero

Set the resulting velocity equation equal to zero and solve for 't'. The value(s) of 't' represent the time when the velocity is zero.
04

Derive the acceleration function

Use the derivative to convert the velocity function into an acceleration function. Differentiate the velocity function \(v(t)\). The derivative of \(v(t)\) will give \(a(t)\). Again apply the power rule.
05

Calculate position and acceleration at zero velocity instances

Substitute the value(s) of 't' found in step 3 into the position function 'x(t)' and acceleration function 'a(t)' to find the position and acceleration of the car when its velocity is zero.
06

Draw the graphs

For part (b) Plot the motion graphs for position-time, velocity-time, and acceleration-time between t = 0 and t = 2 s. For this, you could pick values for 't' between this range and calculate the corresponding 'x', 'v' and 'a' values, then plot these points on their respective graphs.

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Key Concepts

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

Position-Time Graph
When you're looking at a position-time graph, you're essentially tracking how the position of an object changes with respect to time. For instance, with the test car's position given by the equation \(x(t)=2.17 \text{m}+(4.80 \text{m/s}^2)t^2-(0.100 \text{m/s}^6)t^6\), you can plot \(x\) against \(t\) to visualize the car's movement over time. Such a graph helps in understanding the car's motion, whether it's speeding up, slowing down, or coming to a stop.

Real-life Application

Look around and you'll see position-time graphs in action. For example, the GPS tracker in a delivery truck uses this concept to display the vehicle's position over the course of a day.
Velocity-Time Graph
Now let's shift gears to the velocity-time graph. It shows how the velocity of an object changes over time. By differentiating the car's position function, you get the velocity function, which is the car's speed in a specific direction at a certain instant. A flat line on this graph indicates constant speed, while a sloping line indicates acceleration or deceleration.

Typical Scenario

Imagine you're timing your morning jog on a stopwatch鈥攜ou're actually creating a rudimentary velocity-time graph where spikes in your speed (like when you sprint) would show as steep slopes on the graph.
Acceleration-Time Graph
Acceleration-time graphs step it up a notch by displaying acceleration, which is how quickly the velocity of an object is changing. It's the 'rate of change' of velocity. By taking the derivative of the velocity function, you find the car's acceleration at any point in time. Acceleration signifies not just speed changes, but direction shifts too.

Everyday Example

Think of an elevator. The initial jolt you feel when it starts moving鈥攖hat's acceleration. An acceleration-time graph for the elevator's ride would show a spike at the beginning and end (when it starts and stops) and a steadier line while it's in motion between floors.
Motion Equations
Motion equations are the bread and butter of kinematics, giving us the mathematical descriptions of motion. They tie together position, velocity, and acceleration, allowing us to predict where and how fast something will be at any given time. These equations are especially useful when motion involves uniform acceleration or deceleration.

Practical Use

Think about theme park rides. Engineers use motion equations to design roller coasters so that they provide thrills without exceeding safe speed limits at any point during the ride.
Derivative Application in Physics
Derivatives aren't just a math class headache鈥攖hey're super useful in physics, particularly in kinematics. When applied to physics, derivatives give you the rate at which a variable changes. For the test car's movement, you used the derivative to find the velocity and acceleration from the position function. It's all about finding the 'instantaneous' rates, like snapping a speedometer reading at a precise moment.

Impact on Technology

Automatic braking systems in cars are a direct application of derivatives. They continuously calculate the car鈥檚 velocity and how quickly it's changing, which helps to apply the brakes just right to prevent collisions or reduce impact.

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Most popular questions from this chapter

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, \(6.00 \mathrm{~s}\) after it was thrown. What is the speed of the rock just before it reaches the water \(28.0 \mathrm{~m}\) below the point where the rock left your hand? Ignore air resistance.

A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is \(5.0 \mathrm{~m}\) above the surface and has a downward speed of \(0.8 \mathrm{~m} / \mathrm{s}\). With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is \(1.6 \mathrm{~m} / \mathrm{s}^{2}\).

A race car starts from rest and travels east along a straight and level track. For the first \(5.0 \mathrm{~s}\) of the car's motion, the castward component of the car's velocity is given by \(v_{x}(t)=\left(0.860 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\) What is the acceleration of the car when \(v_{x}=12.0 \mathrm{~m} / \mathrm{s} ?\)

A juggler throws a bowling pin straight up with an initial speed of \(8.20 \mathrm{~m} / \mathrm{s}\). How much time elapses until the bowling pin returns to the juggler's hand?

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof \(1.00 \mathrm{~s}\) later. Ignore air resistance. (a) If the height of the building is \(20.0 \mathrm{~m},\) what must the initial speed of the first ball be if both are to hit the ground at the same time? On the same graph, sketch the positions of both balls as a function of time, measured from when the first ball is thrown. Consider the same situation, but now let the initial speed \(v_{0}\) of the first ball be given and treat the height \(h\) of the building as an unknown. (b) What must the height of the building be for both balls to reach the ground at the same time if (i) \(v_{0}\) is \(6.0 \mathrm{~m} / \mathrm{s}\) and (ii) \(v_{0}\) is \(9.5 \mathrm{~m} / \mathrm{s} ?\) (c) If \(v_{0}\) is greater than some value \(v_{\max },\) no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\max }\). The value \(v_{\max }\) has a simple physical interpretation. What is it? (d) If \(u_{0}\) is less than some value \(v_{\min }\), no value of \(h\) exists that allows both balls to hit the ground at the same time. Solve for \(v_{\min }\). The value \(v_{\min }\) also has a simple physical interpretation. What is it?
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