Answer:
1) t_total = 140 min , v = 126.7 m / min , 2) v = 0.628 m / s ,
3) v = 0.592 m / s , θ = -39.3 º
Step-by-step explanation:
Here we have some short kinematics problems
1) d₁ = 4200 m at T1 = 25 min
d₂ = 4.54 km at t2 = 1 h
d₃ = 9000 m at t3 = 55 min
of the statement the direction of each route is the same
Let's reduce the distances to the SI system
d₂ = 4.54 km (1000 m / 1 km) = 4540 m
t₂ = 1 h (60 min / 1 h) = 60 min
Total time is
t_total = t1 + t2 + t3
t_total = 25 + 60 +55
t_total = 140 min
the distance is
d_total = 4200 + 4540 +9000
d_total = 17740 m
with movement it is in one dimension and all displacement is in the same direction, distance equals displacement.
Speed is
v = 17740/140
v = 126.7 m / min
in the direction of travel
Speed is the modulus of speed
| v | = 126.7 m / min
2) the angular velocity is
w = θ / t
In this exercise
t = 2 min (60 s / 1min) = 120 s
θ = 30 rev (2π rad / 1 rev) = 60π rad
w = 60π / 120
w = 1.57 rad / s
linear velocity is
v = w r
v = 1.57 0.40
v = 0.628 m / s
velocity is the speed tangential to the trajectory
3) the data for this exercise are
distance = (number of steps) (distance of one step)
d₁ = 500 0.40 m = 200 m West
d₂ = 400 0.35 m = 140 m North
d₃ = 300 0.30 m = 90 m East
in a total time of t = 2 min and 120 s
t = 2 60 + 120 = 240 s
the distance traveled is
x axis we assume the direction towards + x (East) positive
x_total = d₃ -d₁
x_total = 90-200 = -110 m
y axis
y_total = d₂
y_total = 140 m
we use the Pythagorean theorem to find the distance0
D = √ (x_total² + y_total²)
D = √ (110² + 90²)
D = 142.13 m
The speed is
v = D / t
v = 142.13 / 240
v = 0.592 m / s
speed is
v = 0592 m / s
tan θ = Y / X
θ = tan -1 y / x
θ = tan -1 (90 / (- 110))
θ = -39.3 º
4) This angle is measured on the positive side of the x axis in a clockwise direction
θ = 1.5 rev (2π rad / 1 rev) = 3π rad
t = 90 s
the angular velocity is
w = θ / t
w = 3πi / 90
w = 0.1047 rad / s
the speed is
v = w R
v = 0.14047 250
v = 26.17 m / s
velocity is tangent to path