A particle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗
=(−1.90 m/s2)xˆ+(3.20 m/s2)yˆa→=(−1.90 m/s2)x^+(3.20 m/s2)y^ . (a) How far does the particle move in the x direction before turning around? (b) What is the particle’s velocity at this time? (c) Plot the particle’s position at t=0.500 st=0.500 s , 1.00 s, 1.50 s, and 2.00 s. Use these results to sketch position versus time for the particle.
1 answer:
Answer:
The distance before stopping is 1.52 m,
velocity is 4.0 m/s on the y-axis
Explanation:
The particle’s motion is two-dimensional due to acceleration along both the x and y axes; each axis can be addressed independently for calculations.
a) At the moment the particle starts to reverse, its velocity should be zero (Vfx = 0)
Vfₓ = V₀ₓ + aₓ t
t = - V₀ₓ/aₓ
t = - 2.4/(-1.9)
t= 1.26 s
When the particle stops, we calculate its position
X1 = V₀ₓ t + ½ aₓ t²
X1= 2.4 1.26 + ½ (-1.9) 1.26²
X1= 1.52 m
At this point, the particle begins returning.
b) The velocity comprises both x and y components.
For the x section, Vₓ = 0 m/s indicates a halt, but the y component retains a velocity
Vfy= Voy + ay t
Vfy= 0 + 3.2 1.26
Vfy = 4.0 m/s
Thus, the velocity reads as
V = (0 x^ + 4.0 y^) m/s
c) To graph the motion, we create a table listing position x and y at given time intervals; let's begin the calculations for equations
X = V₀ₓ t+ ½ aₓ t²
Y = Voy t + ½ ay t²
X= 2.4 t + ½ (-1.9) t²
Y= 0 + ½ 3.2 t²
X= 2.4 t – 0.95 t²
Y= 1.6 t²
With these equations, we construct two graphs for position against time, one for the x-axis and another for the y-axis
Chart for graphing
Time (s) x(m) y(m)
0 0 0
0.5 0.960 0.4
1 1.45 1.6
1.50 1.46 3.6
2.00 1.00 6.4
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Answer:
130 m/s (to two significant figures)
Explanation:
In projectile motion, the launching velocity and launch angle help to determine both the horizontal and vertical velocity components.
u represents the initial projectile velocity = 150 m/s
uₓ = u cos θ = 150 cos 30° = 129.9 m/s
uᵧ = u sin θ = 150 sin 30° = 75.0 m/s
A projectile's motion can be viewed as made up of independent vertical and horizontal elements.
The vertical motion is affected by gravitational acceleration (which pulls down on the projectile), altering the vertical velocity component due to this acting force.
Conversely, there is no acting force in the horizontal direction, which means the horizontal component maintains a steady velocity throughout the projectile's flight.
Thus, at t = 4 s, the horizontal component of the projectile's speed remains equal to the initial horizontal velocity component.
At t = 4 s, the horizontal component of velocity is uₓ = u cos θ = 150 cos 30° = 129.9 m/s ≈ 130 m/s
Answer
Data provided:
mass of the block = 200 g = 0.2 Kg
Velocity at A = 0 m/s
Velocity at B = 8 m/s
distance of slide = 10 m
height of the block = 4 m
calculation for the block's potential energy
P = m g h
P = 0.2 x 9.8 x 4
P = 7.84 J
kinetic energy calculated as
Work done = P - KE
work = 7.84 - 6.14
work = 1.7 J
b) using the formula v² = u² + 2 a s
0 = 8² - 2 x a x 10
a = 3.2 m/s²
ma - μ mg = 0