Question

Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a speed of 2.0 m/s. Their dog, Spot, starts by Arthur's side at the same time and runs back and forth between them at 5.0 m/s. By the time Arthur and Betty meet, what distance has Spot run?

Answer 1

Let t be the time they both meet.  When they meet they both have covered the distance between them.

Arthur with 3 m/s would have covered 3t m
Betty with 2 m/s would have covered 2t m

3t + 2t = 100
5t = 100
t = 100/5
t = 20 seconds.

Spot who is helplessly runing back and forth at 5 m/s would have done this for 20 seconds.

This distance covered by spot =  5m/s * 20s = 100m

Answer 2

The distance Spot has traveled is 100 m

Further explanation

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem!

Given:

distance = d = 100 m

Arthur's speed = v₁ = 3.0 m/s

Betty's speed = v₂ = 2.0 m/s

Spot's speed = v = 5.0 m/s

Unknown:

Spot's distance = s = ?

Solution:

At first we will find time when the two of them meet.

$\text {Arthur's distance + Betty's Distance} = 100 ~ m$

$v_1t + v_2t = 100$

$3t + 2t = 100$

$5t = 100$

$t = 100 \div 5$

$\boxed {t = 20 ~ seconds}$

Because Spot runs back and forth until these two people meet, then:

$\text {Spot's Distance} = vt$

$\text {Spot's Distance} = 5(20)$

$\large {\boxed {\text {Spot's Distance} = 100 ~ m} }$

Learn more

• Velocity of Runner : brainly.com/question/3813437
• Kinetic Energy : brainly.com/question/692781
• Acceleration : brainly.com/question/2283922
• The Speed of Car : brainly.com/question/568302

Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle