1. Find the banking angle of the railway track with a radius of curvature of 1500m. If the train's maximum speed is 15m/s. If the distance between the two tracks is 1.8m Calculate the elevation of the outer track over the inner track.
2. Given displacement vectors S₁=(-2i +3j)m, S₂=(3i-2j)m, and S₃= (-2i+3j)m. Find the magnitude and direction of the resultant displacement. S=S₁ S₂ + S3
3. A particle moving at 25 m/s in a straight line slows uniformly at a rate of 2 m/s every second. In an interval of 10s and find
(a) the acceleration, (b) the final velocity, (c) the distance moved.
4. On its way along a straight line, a car starts from rest and accelerated at a rate of 2.5 m/s² for 10s. It maintained its speed for further 5 s and decelerated by 0.5m/s² to come to rest. What is the time taken for the entire motion? How far did it travel? Draw the position vs time, and velocity vs time graph of the motion of the car
5. Three force vectors A, B and C are acting at a point. If A=4i-3j.B=2i+j and A-B-C-0. Find
a/ the magnitude and direction of vector C and b/a unit vector along C
6 . The initial position vector of a particle on the xy plane is (5.2) and its final position vector is (-2,-3). If distance is measured in centimeter,
a) what is the displacement of the particle?
b) What is the magnitude and direction of the displacement?
1. The banking angle of the railway track can be found using the equation:
tan(θ) = (v^2)/(g*r)
where θ is the banking angle, v is the maximum speed of the train, g is the acceleration due to gravity, and r is the radius of curvature.
Plugging in the given values:
v = 15 m/s
r = 1500 m
g = 9.8 m/s^2
tan(θ) = (15^2)/(9.8*1500)
tan(θ) = 0.15
θ = arctan(0.15)
Using a scientific calculator, the banking angle is approximately 8.53 degrees.
To calculate the elevation of the outer track over the inner track, we can use trigonometry. Since the distance between the two tracks is the horizontal distance between the two edges of the track, we can use the tangent function:
tan(θ) = h/d
where h is the elevation difference and d is the distance between the tracks.
Plugging in the given values:
tan(8.53) = h/1.8
h = 1.8 * tan(8.53)
Using a scientific calculator, the elevation of the outer track over the inner track is approximately 0.28 meters.
2. To find the magnitude and direction of the resultant displacement, we can use the vector addition equation:
S = S₁ + S₂ + S₃
S = (-2i + 3j) + (3i - 2j) + (-2i + 3j)
S = -i + i + 3j - 2j + 3j
S = 2j + 4j
S = 6j
The magnitude of the resultant displacement is |S| = |6j| = 6 meters.
The direction of the resultant displacement can be found using the equation:
θ = arctan(Sy/Sx)
where θ is the direction angle, Sy is the y-component of the resultant displacement, and Sx is the x-component of the resultant displacement.
In this case, Sy = 6 and Sx = 0 (since there is no x-component).
θ = arctan(6/0) = 90 degrees
Therefore, the direction of the resultant displacement is straight up, vertically upward.
3. (a) The acceleration can be found using the equation:
a = (v - u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time interval.
Plugging in the given values:
v = 0 m/s (since the particle comes to a stop)
u = 25 m/s
t = 10 s
a = (0 - 25) / 10 = -2.5 m/s²
The acceleration is -2.5 m/s².
(b) The final velocity can be found using the equation:
v = u + at
Plugging in the given values:
u = 25 m/s
a = -2 m/s²
t = 10 s
v = 25 + (-2) * 10 = 25 - 20 = 5 m/s
The final velocity is 5 m/s.
(c) The distance moved can be found using the equation:
s = ut + (1/2)at²
Plugging in the given values:
u = 25 m/s
a = -2 m/s²
t = 10 s
s = 25 * 10 + (1/2) * (-2) * (10)^2 = 250 - 100 = 150 m
The distance moved is 150 m.
4. To find the time taken for the entire motion, we need to break the motion into three parts: acceleration, constant speed, and deceleration.
(a) Acceleration:
Using the equation for constant acceleration:
v = u + at
Plugging in the given values:
u = 0 m/s (since the car starts from rest)
a = 2.5 m/s² (acceleration rate)
t₁ = 10 s
v₁ = 0 + 2.5 * 10 = 25 m/s
(b) Constant speed:
The car maintains a constant speed for 5 seconds, so the distance traveled can be calculated using:
s = vt
Plugging in the given values:
v₂ = 25 m/s (constant speed)
t₂ = 5 s
s₂ = 25 * 5 = 125 m
(c) Deceleration:
Using the equation for constant deceleration:
v = u + at
Plugging in the given values:
u = 25 m/s (constant speed)
a = -0.5 m/s² (deceleration rate)
t₃ = unknown
0 = 25 - 0.5 * t₃
0.5 * t₃ = 25
t₃ = 50 s
The time taken for the entire motion is t = t₁ + t₂ + t₃ = 10 + 5 + 50 = 65 s.
To find the total distance traveled, we sum the distances from each part of the motion:
s = s₁ + s₂ + s₃
s = 150 + 125 + 0 = 275 m
The car travels a total distance of 275 m.
For the position vs time graph, the motion can be split into three segments: acceleration, constant speed, and deceleration. The graph for each segment would be a straight line at different slopes. At t = 0 to t = 10, the graph would have a positive slope (acceleration), at t = 10 to t = 15, the graph would be horizontal (constant speed), and at t = 15 to t = 65, the graph would have a negative slope (deceleration).
The velocity vs time graph would have a positive slope from t = 0 to t = 10 (acceleration), a horizontal line at 25 m/s from t = 10 to t = 15 (constant speed), and a negative slope from t = 15 to t = 65 (deceleration).
5. To find the magnitude and direction of vector C, we can rearrange the equation A - B - C = 0:
C = A - B
C = (4i - 3j) - (2i + j)
C = 4i - 3j - 2i - j
C = 2i - 4j
The magnitude of vector C is |C| = sqrt((2^2) + (-4^2)) = sqrt(4 + 16) = sqrt(20) = 2sqrt(5).
The direction of vector C can be found using the equation:
θ = arctan(Cy/Cx)
where θ is the direction angle, Cy is the y-component of vector C, and Cx is the x-component of vector C.
In this case, Cy = -4 and Cx = 2.
θ = arctan(-4/2) = arctan(-2) = -63.43 degrees
Therefore, the magnitude of vector C is 2sqrt(5) and the direction is -63.43 degrees with respect to the positive x-axis.
To find a unit vector along C, we divide vector C by its magnitude:
uC = C/|C|
uC = (2i - 4j)/2sqrt(5)
Simplifying, we get:
uC = (i - 2j)/sqrt(5)
Therefore, a unit vector along C is (1/sqrt(5))i - (2/sqrt(5))j.
6. (a) The displacement of the particle can be found by subtracting the initial position vector from the final position vector:
Displacement = Final position - Initial position
Displacement = (-2i, -3j) - (5i, 0j)
Displacement = (-2i - 5i, -3j - 0j)
Displacement = -7i, -3j
The displacement of the particle is -7i - 3j.
(b) The magnitude of the displacement can be found using the equation:
Magnitude of displacement = sqrt((Displacement in x)^2 + (Displacement in y)^2)
Magnitude of displacement = sqrt((-7)^2 + (-3)^2)
Magnitude of displacement = sqrt(49 + 9)
Magnitude of displacement = sqrt(58)
The magnitude of the displacement is sqrt(58) cm.
The direction of the displacement can be found using the equation:
Direction of displacement = arctan(Displacement in y / Displacement in x)
Direction of displacement = arctan(-3 / -7)
Direction of displacement = arctan(3/7)
Therefore, the magnitude of the displacement is sqrt(58) cm and the direction is arctan(3/7) with respect to the positive x-axis.