Figure 2-45 shows a simple device for measuring your reaction time. It consists of a cardboard strip marked with a scale and two large dots. A friend holds the strip vertically, with thumb and forefinger at the dot on the right in Fig. 2-45. You then position your thumb and forefinger at the other dot (on the left in Fig. 2-45), being careful not to touch the strip. Your friend releases the strip, and you try to pinch it as soon as possible after you see it begin to fall. The mark at the place where you pinch the strip gives your reaction time. (a) How far from the lower dot should you place the 50.0 m smark? How much higher should you place the marks for (b) 100, (c) 150, (d) 200, and (e) 250 ms? (For example, should the 100 msmarker be 2 times as far from the dot as the 50msmarker? If so, give an answer of 2 times. Can you find any pattern in the answers?

Short Answer

Expert verified
  1. The position of the 50.0 ms mark from the lower dot, y1=1.23cm
  2. The position of the 100 ms mark from the lower dot is 4.92cm
  3. The position of the 150 ms mark from the lower dot is 11.07 cm
  4. The position of the 200 ms mark from the lower dot is 19.68 cm
  5. The position of the 250 ms mark from the lower dot is 30.75 cm

Step by step solution

01

Kinematic Equations of Motion and Concept of Free Fall

The problem deals with the kinematic equation of motion. Also, it involves the concept of free fall. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion. The situation of a body moving freely in any direction in the presence of gravity is referred to as free fall.

Formula:

The displacement in kinematic equation is given by,

y=v0t+12at2

Where, v0is the initial velocity a is acceleration and t is time.

02

a): Calculations for distance of the lower dot from 50 m/s mark

According to the kinematic equations,

y=v0t+12at20-y1=0-129.80.052y1=4.90.0025y1=0.0123m~1.23cm

Hence, the distance of lower dot from mark would be 1.23 cm

03

b): Calculations for distance of the lower dot from 100 ms mark

y=v0t+12at20-y2=0-129.80.12y2=4.90.01y2=0.049m=4y1

Hence, the distance of lower dot from 100 ms mark would be4y1

04

c): Calculations for distance of the lower dot from 150 ms mark

y=v0t+12at20-y3=0-129.80.152y3=4.90.0225y3=0.11m=9y1

Hence, the distance of lower dot from 150 ms mark would be9y1

05

d): Calculations for distance of the lower dot from 200 ms mark

y=v0t+12at20-y4=0-129.80.22y4=4.90.04y4=0.196m=16y1

Hence, the distance of lower dot from 200 ms mark would be16y1

06

e): Calculations for distance of the lower dot from 250 ms mark

y=v0t+12at20-y5=0-129.80.252y5=4.90.0625y5=0.306m=25y1

Hence, the distance of lower dot from 250 ms mark would be25y1

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The 1992 world speed record for a bicycle (human powered vehicle) was set by Chris Huber. His time through the measured 200 m stretch was a sizzling 6.509 sec , at which he commented “Cogito ergo zoom!” (I think, therefore I go fast). In 2001, Sam Whittingham beat Huber’s record by 19.0 km/h. What was Whittingham’s time through the 200 m?

A stone is thrown vertically upward. On its way up it passes point A with speed v, and point B ,3.00 m ,higher than A, with speed12vCalculate (a) the speed v and (b) the maximum height reached by the stone above point B.

The position function x(t)of a particle moving along an x axis is x=4-6t2, with x in metres and t in seconds. (a) At what time does the particle momentarily stop? (b) Where does the particle momentarily stop? At what (c) negative time and (d) positive time does the particle pass through the origin? (e) Graph x vs t for range -5sec to +5sec. (f) To shift the curve rightward on the graph, should we include the term role="math" localid="1657352173540" +20t or -20t in ? (g) Does that inclusion increase or decrease the value of x at which the particle momentarily stops?

A ball of moist clay falls 15.0m to the ground. It is in contact with the ground for 20.0m sbefore stopping. (a) What is the magnitude of average acceleration of the ball during the time it is in contact with the ground? (treat the ball as a particle). (b) Is the average acceleration up or down?

You are to drive 300km to an interview. The interview is at 11.15 AM. You plan to drive at 100 km/h , so you leave at 8 AM, to allow some extra time. You drive at that speed for the first 100 km, but then construction work forces you to slow to 40 km/h for 40km. What would be the least speed needed for the rest of the trip to arrive in time for the interview?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free