The coin will normally land on our hand.
Answer:
Whats the question/word problem or where is the graph (if included) representing this problem?
Answer:
7.15 m/s
Explanation:
We use a frame of reference in which the origin is at the point where the trucck passed the car and that moment is t=0. The X axis of the frame of reference is in the direction the vehicles move.
The truck moves at constant speed, we can use the equation for position under constant speed:
Xt = X0 + v*t
The car is accelerating with constant acceleration, we can use this equation
Xc = X0 + V0*t + 1/2*a*t^2
We know that both vehicles will meet again at x = 578
Replacing this in the equation of the truck:
578 = 24 * t
We get the time when the car passes the truck
t = 578 / 24 = 24.08 s
Before replacing the values on the car equation, we rearrange it:
Xc = X0 + V0*t + 1/2*a*t^2
V0*t = Xc - 1/2*a*t^2
V0 = (Xc - 1/2*a*t^2)/t
Now we replace
V0 = (578 - 1/2*1.4*24.08^2) / 24.08 = 7.15 m/s
Answer:Given:
Initial speed of fugitive, v0 = 0 m/s
Final speed, vf = 6.1 m/s
acceleration, a = 1.4 m/s^2
Speed of train, v = 5.0 m/s
Solution:
t = (vf-v0)/a
t = (6.1-0)/1.4
t =4.36 s
Distance traveled by train, x_T =v*t
x_T =5*4.36 = 21.8 m
Distance travelled by fugitive, x_f = v0*t+1/2at^2
x_f = 0*4.36+1/2*1.4*4.36^2
x_f =13.31 m
5*t = v(t-4.36)+x_f
5*t=6.1*(t-4.36)+13.31
solve for t, we get
t = 12.08 s
The fugitive takes 12.08 s to catch up to the empty box car.
Distance traveled to reach the box car is
X_T = v*t
X_T = 5*12.08 s
X_T = 60.4 m
Explanation:
