/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A sprinter runs a \(100 \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A sprinter runs a \(100 \mathrm{~m}\) dash in \(12.0 \mathrm{~s}\). She starts from rest with a constant acceleration \(a_{x}\) for \(3.0 \mathrm{~s}\) and then runs with constant speed for the remainder of the race. What is the value of \(a_{x} ?\)

Short Answer

Expert verified
To find out the acceleration during the first 3 seconds, use the formula for distance in terms of initial velocity, acceleration and time. Finding out the distance covered during the phase of constant speed allows to calculate the speed during this phase. From this, one can calculate the acceleration in the initial 3 seconds.

Step by step solution

01

Determine distance covered during acceleration

The initial velocity (\(v_{i}\)) of the sprinter is 0, and she accelerates for 3 seconds. Thus, the distance (\(d_{a}\)) covered during this phase can be found out using this formula: \(d_{a}=\frac{1}{2}a_{x}t^{2}\) where \(t\) is time and \(a_{x}\) is acceleration.
02

Determine distance covered during constant speed

The sprinter covers the remaining distance (\(d_{c}\)) of the race with constant speed. This is computed by subtracting the distance covered during acceleration from the total distance. So, \(d_{c} = 100 - d_{a}\)
03

Determine time taken at constant speed

The sprinter runs at constant speed for the remaining time, calculated by subtracting the initial 3 seconds from the total time. So, \(t_{c} = 12 - 3 = 9s \)
04

Compute speed at constant velocity

The speed (\(v_{c}\)) at constant velocity can be calculated using this formula: \(v_{c} = \frac{d_{c}}{t_{c}}\)
05

Calculate the acceleration

The acceleration \(a_{x}\) can be computed using the formula: \(a_{x} = \frac{v_{c}}{t}\). Substitute values of \(v_{c}\) and \(t_{3s}\) to get the acceleration \(a_{x}\).

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Key Concepts

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

Constant Acceleration
Constant acceleration refers to a scenario where an object's velocity changes at a steady rate over time. In such instances, the acceleration - the rate of change of velocity - remains the same throughout the given period. For learners tackling kinematics problems, understanding this concept is crucial as it dictates how objects move when influenced by constant forces.

For example, when a sprinter starts off from rest, as in our exercise, her acceleration is the same over the first 3 seconds of her sprint. This consistent increase in speed is described using equations of kinematics that are specifically derived for constant acceleration conditions. Knowing this helps in predicting how fast the sprinter will be running at any moment within those 3 seconds.
Distance-Time Relationship
The distance-time relationship is a foundational concept in kinematics that describes how an object's displacement is connected to the time elapsed. This relationship can be visualized with a distance-time graph, where the slope of the line indicates the object's speed or velocity.

In our case study of the sprinter, we apply this principle to calculate how far she travels during the acceleration phase. Using the formula \( d_a = \frac{1}{2}a_xt^2 \), we employ the known time duration and the acceleration to find the distance. The distance-time relationship, especially when depicted in graphical form, offers an intuitive understanding of the motion that textual explanation alone may not convey as effectively.
Constant Velocity
After experiencing constant acceleration, the sprinter continues the race at a constant velocity. Constant velocity motion implies that there's no further change in speed; speed and direction both remain the same. This uniform motion directly contrasts with the previous phase of acceleration and complicates the overall motion analysis of the sprinter.

For kinematics problems, recognizing when an object transitions from an accelerated motion to constant velocity is pivotal. In our exercise, once the sprinter attains her top speed at the end of the acceleration phase, her distance covered per second remains consistent. This is calculated using the distance-time formula \( v_c = \frac{d_c}{t_c} \) for constant velocity, showcasing a linear relationship between distance and time.
Uniformly Accelerated Motion
Uniformly accelerated motion denotes a scenario where an object is speeding up or slowing down at a constant rate, which our sprinter experiences during the first 3 seconds of her dash. It's essential to differentiate this from constant velocity because, in this phase, the speed isn't steady but is increasing uniformly.

Creating equations based on the initial conditions and the definitions of uniformly accelerated motion allows us to solve for unknowns such as the sprinter’s acceleration and distance covered. The formula \( a_x = \frac{v_c}{t} \) provided in our exercise solution, derives from these uniform acceleration principles. By thoroughly understanding uniformly accelerated motion, learners can tackle a variety of kinematics problems involving objects in different forms of motion.

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Most popular questions from this chapter

The acceleration of a bus is given by \(a_{x}(t)=\alpha t,\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s}\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?(\mathrm{c})\) Sketch \(a_{y}-t\) \(v_{y}-t,\) and \(x-t\) graphs for the motion.

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of \(20 \mathrm{~m} / \mathrm{s}\) (45 mi/h) when it reaches the end of the 120 -m-long ramp. (a) What is the acceleration of the car? (b) How much time does it take the car to travel the length of the ramp? (c) The traffic on the freeway is moving at a constant speed of \(20 \mathrm{~m} / \mathrm{s}\). What distance does the traffic travel while the car is moving the length of the ramp?

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?

Starting from a pillar, you run \(200 \mathrm{~m}\) east (the \(+x\) -direction) at an average speed of \(5.0 \mathrm{~m} / \mathrm{s}\) and then run \(280 \mathrm{~m}\) west at an average speed of \(4.0 \mathrm{~m} / \mathrm{s}\) to a post. Calculate (a) your average speed from pillar to post and (b) your average velocity from pillar to post.

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth's surface and is to reach a maximum height of \(960 \mathrm{~m}\) above the earth's surface. The rocket's engines give the rocket an upward acceleration of \(16.0 \mathrm{~m} / \mathrm{s}^{2}\) during the time \(T\) that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance. What must be the value of \(T\) in order for the rocket to reach the required altitude?

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