Respuesta :
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
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
The distance Spot has traveled is 100 m
Further explanation
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
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.
[tex]\text {Arthur's distance + Betty's Distance} = 100 ~ m[/tex]
[tex]v_1t + v_2t = 100[/tex]
[tex]3t + 2t = 100[/tex]
[tex]5t = 100[/tex]
[tex]t = 100 \div 5[/tex]
[tex]\boxed {t = 20 ~ seconds}[/tex]
Because Spot runs back and forth until these two people meet, then:
[tex]\text {Spot's Distance} = vt[/tex]
[tex]\text {Spot's Distance} = 5(20)[/tex]
[tex]\large {\boxed {\text {Spot's Distance} = 100 ~ m} }[/tex]
Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
Answer details
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle
