<h3>
</h3><h3>Given</h3>
v = 20m\s
a = 3m\s^2
t = 4sec
Firstly we have to find u
a = 
3m\s =
12m\s = 20 - u
20 - u = 12m\s
- u = -8
u = 8
Now we can easily find distance by using second equation of motion
s = ut + 1\2 at^2
s = 8(4) + 1\2(3)(16)
s = 32 + 24
s = 56
So distance is 56 m\s hope it helps
Answer:
-1.03 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity. The S. I unit of acceleration is m/s².
Mathematically, acceleration is expressed as
a = (v-u)/t ........................ Equation 1
Where a = acceleration, v = final velocity, u = initial velocity, t = time.
Given: u = 13.60 m/s, v = 7.20 m/s t = 6.2 s.
Substituting into equation 2
a = (7.20-13.60)/6.2
a = -6.4/6.2
a = -1.03 m/s²
Note: a is negative because, the hockey puck is decelerating.
Hence the average acceleration = -1.03 m/s²
A: Human Body
C is wrong because they don’t have the tools to test it on another planet
No, because sometimes you have to stop at stop signs and stop lights.
