1)A hiker's trip consists of three segments. Path A is 7.0 km long heading 60.0° north of east. Path B is 6.0 km long in a direction due east. Path C is 5.0 km long heading 315° counterclockwise from east.
A) Graphically add the hiker's displacements in the order A, B, C.
Magnitude of displacement ?
Direction of displacement ?
B) Graphically add the hiker's displacements in the order C, B, A.
Magnitude of displacement ?
Direction of displacement?
C) What can you conclude about the resulting displacements?
2)When you see a traffic light turn red, you apply the brakes until you come to a stop. Suppose your initial speed was 11.8 m/s, and you come to rest in 36.0 m. How much time does this take? Assume constant deceleration.
A) Graphically add the hiker's displacements in the order A, B, C.
Magnitude of displacement ?
Direction of displacement ?
B) Graphically add the hiker's displacements in the order C, B, A.
Magnitude of displacement ?
Direction of displacement?
C) What can you conclude about the resulting displacements?
2)When you see a traffic light turn red, you apply the brakes until you come to a stop. Suppose your initial speed was 11.8 m/s, and you come to rest in 36.0 m. How much time does this take? Assume constant deceleration.
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Trip section A can be broken down into N and E components
N = 7sin60 = 6.06 km
E = 7cos60 = 3.5 km
Section B has only one direction and it is E
E = 6 km
Section C can be broken down into N and E components where South is negative North
N = 5sin315 = -3.54 km
E = 5cos315 = 3.54 km
A) Graphically you would draw a line from the origin 7 units long along a direction of 60 degrees north of east. From the end of this line head directly east for 6 units. From the end of the second line segment, draw a line 5 units long along a direction of 45 degrees south of east, which equals 360 - 45 = 315 degrees CCW from east.
The total displacements by components will be
N = 6.06 - 3.54 = 2.52 km
E = 3.5 + 6 + 3.54 = 13.04 km
so the total displacement is
(2.52^2 + 13.04^2)^0.5 = 13.28 km
and the direction is inv tan (2.52/13.04) or 10.9 degrees North of East
B) Graphically adding them in C,B,A order, from the origin draw a line 5 units long at a -45 degree angle. From the end of this line draw a line 6 units long directly east. From the end of the second segment, draw a line 7 units long along a direction 60 degrees north of east.
c) the total direction and magnitude of the resultant will be identical to those found in answer A) because all the components are identical and the order of addition is not relevant to the result.
2 distance under constant acceleration can be found using the equation
d = (1/2)a*t^2
velocity is acceleration multiplied by time
v = a*t
using these two equations, we can sub the second into the first giving us
d = (1/2)a*t*t
d = (1/2)v*t
so that
t = 2d/v
t = 2(36)/11.8
t = 6.1 seconds
N = 7sin60 = 6.06 km
E = 7cos60 = 3.5 km
Section B has only one direction and it is E
E = 6 km
Section C can be broken down into N and E components where South is negative North
N = 5sin315 = -3.54 km
E = 5cos315 = 3.54 km
A) Graphically you would draw a line from the origin 7 units long along a direction of 60 degrees north of east. From the end of this line head directly east for 6 units. From the end of the second line segment, draw a line 5 units long along a direction of 45 degrees south of east, which equals 360 - 45 = 315 degrees CCW from east.
The total displacements by components will be
N = 6.06 - 3.54 = 2.52 km
E = 3.5 + 6 + 3.54 = 13.04 km
so the total displacement is
(2.52^2 + 13.04^2)^0.5 = 13.28 km
and the direction is inv tan (2.52/13.04) or 10.9 degrees North of East
B) Graphically adding them in C,B,A order, from the origin draw a line 5 units long at a -45 degree angle. From the end of this line draw a line 6 units long directly east. From the end of the second segment, draw a line 7 units long along a direction 60 degrees north of east.
c) the total direction and magnitude of the resultant will be identical to those found in answer A) because all the components are identical and the order of addition is not relevant to the result.
2 distance under constant acceleration can be found using the equation
d = (1/2)a*t^2
velocity is acceleration multiplied by time
v = a*t
using these two equations, we can sub the second into the first giving us
d = (1/2)a*t*t
d = (1/2)v*t
so that
t = 2d/v
t = 2(36)/11.8
t = 6.1 seconds