velocity of the physics instructor with respect to bus
acceleration of the bus is given as
acceleration of instructor with respect to bus is given as
now the maximum distance that instructor will move with respect to bus is given as
so the position of the instructor with respect to door is exceed by
so it will be moved maximum by 3 m distance
So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2
Given:
Velocity: 0.5 mile/minute
Time: 12 minute
Now we know that speed and velocity have the same magnitude. Hence speed=velocity=0.5 mile/min
Substituting the given values in the above formula we get
Distance = 0.5 x 12= 6 miles
