Question

A typical running track is an oval with 74-m-diameter half circles at each end. a runner going once around the track covers a distance of 400 m . suppose a runner, moving at a constant speed, goes once around the track in 1 min 40 s. what is her centripetal acceleration during the turn at each end of the track?

Answer 1

The centripetal acceleration of runner at each end of track is $\boxed{0.432{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

Explanation:

It is given that the diameter of oval track that consists of two half circles at each end is $74{\text{ m}}$ .

The distance runner runs once around the track is $400{\text{ m}}$ . The runner covers one round of track in $1{\text{ m}}$  and $40{\text{ s}}$ .

Our aim is obtain the value of centripetal acceleration for runner at each track.

The expression for centripetal acceleration is shown below.

${a_c}=\frac{{{v^2}}}{R}$                                                                          …… (1)

Here, $v$  is the speed of the runner and $R$  is the radius of track.

The speed of the runner is the ratio of distance she takes to cover the track to the time taken by her.

The time in seconds is calculated as,

$\begin{aligned}t&=60{\text{ s}}+40{\text{ s}}\\&=100{\text{ s}}\\\end{aligned}$

The distance covered is $400{\text{ m}}$ , so the speed calculated is shown below.

$\begin{aligned}{\text{speed}}&=\frac{{{\text{distance}}}}{{{\text{time}}}}\\&=\frac{{400{\text{ m}}}}{{100{\text{ s}}}}\\&=4{\text{ }}{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The radius of track calculated is shown below.

$\begin{aligned}R&=\frac{d}{2}\\&=\frac{{74{\text{ m}}}}{2}\\&=37{\text{ m}}\\\end{aligned}$

Substitute $4$  for $v$  and $37$  for $R$  in equation (1) to obtain the centripetal acceleration ${a_c}$ .

$\begin{aligned}a_{c}&=\frac{(4\text{ m/s})^{2}}{37\text{ m}}\\&=\frac{16}{37}\text{ m/s}^{2}\\&=0.432\text{ m/s}^{2}\end{aligned}$

Therefore, the centripetal acceleration ${a_c}$  is $0.432{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}$ .

Thus, the centripetal acceleration of runner at each end of track is $\boxed{0.432{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}$ .

Learn More:

1. Centripetal Force brainly.com/question/6372960

2. Velocity and Momentum brainly.com/question/11896510

3. Centripetal force on orbit brainly.com/question/7420923

Answer Details:

Grade: High School

Subject: Physics

Chapter: Circular Motion

Keywords:

Track, running, diameter, circle, runner, distance, moves, constant, speed, centripetal, acceleration, force, motion, uniform, radius, velocity and oval.

Answer 2

Answer:

Her centripetal acceleration during the turn at each end of the track is $0.432\, \frac{m}{s^{2}}$

Step-by-step explanation:

Total distance covered in one round , D= 400 m

Time taken to cover one round , T = 1 min 40 s = 100 sec

Speed of runner , $V=\frac{D}{T}=\frac{400}{100}\, \frac{m}{s}= 4.0\, \frac{m}{s}$

Now centripetal acceleration is given by

$a_c=\frac{V^{2}}{R}$

where $V= 4.0\frac{m}{s}$

$Radius, R= \frac{Diameter}{2}=\frac{74}{2}m=37\, m$

$\therefore a_c=\frac{4^{2}}{37}\, \frac{m}{s^{2}}=0.432\, \frac{m}{s^{2}}$

Thus her centripetal acceleration during the turn at each end of the track is $0.432\, \frac{m}{s^{2}}$