I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
Well i think the answer is impossible to find because there is no picture
Answer:
The rate of the boat in still water is 44 mph and the rate of the current is 4 mph
Explanation:
x = the rate of the boat in still water
y = the rate of the current.
Distance travelled = 120 mi
Time taken upstream = 3 hr
Time taken downstream = 2.5 hr
Speed = Distance / Time
Speed upstream
Speed downstream
Adding both the equations
The rate of the boat in still water is <u>44 mph</u> and the rate of the current is <u>4 mph</u>
How can one explain<span> and predict the </span>interactions between objects<span> and within a system of </span>objects<span>? ... through </span>electric or magnetic fields to illustrate the forces between objects and the changes in energy of the objects due to the interaction<span>. ... </span>Forces<span> at a distance are </span>explained<span> by fields (gravitational, </span>electric<span>, and magnetic) ...</span>
