Note that this is a position vs time graph.
From A to B, the graph is a straight line with a nonzero slope. This indicates a constant velocity.
From B to C, the graph is a straight line with 0 slope. This indicates a constant position, i.e. the object remains stationary.
From C to D, the graph is a straight line with a nonzero slope. This indicates a constant velocity.
A bowler who always left the same 3 pins standing could be considered a C. Precise bowler as from bowling countless number of times he has observed the same amount of pins knocked down each time.
Answer:
A. The time taken for the car to stop is 3.14 secs
B. The initial velocity is 81.64 ft/s
Explanation:
Data obtained from the question include:
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Final velocity (V) = 0
Time (t) =?
Initial velocity (U) =?
A. Determination of the time taken for the car to stop.
Let us obtain an express for time (t)
Acceleration (a) = Velocity (V)/time(t)
a = V/t
Velocity (V) = distance (s) /time (t)
V = s/t
a = s/t^2
Cross multiply
a x t^2 = s
Divide both side by a
t^2 = s/a
Take the square root of both side
t = √(s/a)
Now we can obtain the time as follow
Acceleration (a) = 26ft/s2
Distance (s) = 256ft
Time (t) =..?
t = √(s/a)
t = √(256/26)
t = 3.14 secs
Therefore, the time taken for the car to stop is 3.14 secs
B. Determination of the initial speed of the car.
V = U + at
Final velocity (V) = 0
Deceleration (a) = –26ft/s2
Time (t) = 3.14 sec
Initial velocity (U) =.?
0 = U – 26x3.14
0 = U – 81.64
Collect like terms
U = 81.64 ft/s
Therefore, the initial velocity is 81.64 ft/s
Ok, so you've got to figure out a force F and you have the speed in which the boxer punches on determinate time and the mass of the sheet of paper.
So based on the formula that says that the Force is equal to the mass multiplied by the acceleration => F=ma.
You look at it and see that you only have mass which is measured on KG so there is no problem.
then you have the acceleration which is measured on meters and is defined by: a = Δv/Δt
So now you can replace the velocity and the time you have there
⇒ a 25m/s / 0.05s
you have computing that ⇒ 50m because the seconds were cancelled out.
and then you plug the meters into the force equation.
F=(0.005kg)(50)
F=0.25N
so the boxer will have a force of 0.25 Newton's.