We will use this equation:
s = 1/2*a*t^2 + v0*t + s0
where:
s = space traveled
a = acceleration
t = time
v0 = initial speed
s0 = initial space
In this case::
v0 = 0
s0 = 0
So our equation will look like that now:
s = 1/2 * a * t^2
let's calculate the acceleration first of all:
a = (vf - vi) / t
where vf is the final speed and vi is the initial speed. t is the time.
a = (25m/s) / 10s = 2.5 m/s^2
Now we can calculate the space:
s = 1/2 * (2.5 m/s^2) * (10s)^2 = 125m
---
Hope it was helpful! Have a great day.
Answer:
The distance covered by puck A before collision is 
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is 
The speed of puck B is 
The distance covered by puck A is mathematically represented as

=> 
The distance covered by puck B is mathematically represented as

=> 
Since the time take before collision is the same

substituting values

=> 
=> 
smaller, because the part of the velocities cancel each other out
The answer should be all of the above
Answer:
correct answer is (c) 15 J
Explanation:
given data
mass m1 = 2 kg
velocity V1 = 5 m/s
mass other = 3 kg
so mass m2 = 2+ 3 kg = 5 kg
solution
we will apply here conservation of momentum:
m1V1 = m2V2 ..........................1
put here value and we get velocity v2
(2.0) × (5.0) = (2.0 + 3.0) × V
solve it we get
10 = 5 × V
2
V2 = 2.0 m/s
so here kinetic energy will be
KE = ½ × m × v²
so
∆KE = ½ × m1 × (v1)² - ½ × m2 × (v2)
²
∆KE = 0.5 × 2 × 25 - 0.5 × 5 × 4
∆KE = 25 - 10
∆KE = 15 J