Concept explainers
A particle initially located at the origin has an acceleration of
(a)
The vector position of the particle at time
Answer to Problem 3P
The vector position of the particle at time
Explanation of Solution
Write the expression for the vector position of the particle.
Here,
Conclusion:
Substitute
Therefore, the vector position of the particle at time
(b)
The velocity of the particle at any time
Answer to Problem 3P
The velocity of the particle at any time
Explanation of Solution
Write the expression for the velocity of the particle at any time
Here,
Conclusion:
Substitute
Therefore, the velocity of the particle at any time
(c)
The coordinates of the particle at
Answer to Problem 3P
The coordinates of the particle at
Explanation of Solution
Write the expression for the vector position of the particle at time
Here,
Conclusion:
Substitute
So compare the above value with (III) to get
Therefore, the coordinates of the particle at
(d)
The speed of the particle at
Answer to Problem 3P
The speed of the particle at
Explanation of Solution
Write the expression for the speed of the particle at time
Write the final speed of the particle.
Here,
Conclusion:
Substitute
Substitute
Therefore, the speed of the particle at
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Chapter 3 Solutions
Principles of Physics: A Calculus-Based Text
- At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of vi=(3.00i2.00j)m/s and is at the origin. At t = 3.00 s, the particles velocity is vf=(9.00i+7.00j)m/s. Find (a) the acceleration of the particle and (b) its coordinates at any time t.arrow_forwardFrom the window of a building, a ball is tossed from a height y0 above the ground with an initial velocity of 8.60 m/s and angle of 18.0° below the horizontal. It strikes the ground 5.00 s later. (a) If the base of the building is taken to be the origin of the coordinates, with upward the positive y-direction, what are the initial coordinates of the ball? (Use the following as necessary: y0.) xi = 0 yi = γ0 (b) With the positive x-direction chosen to be out the window, find the x- and y-components of the initial velocity. vi,x = m/s vi,y = m/s (c) Find the equations for the x- and y- components of the position as functions of time. (Use the following as necessary: y0 and t. Let the variable t be measured in seconds.) x = m y = m (d) How far horizontally from the base of the building does the ball strike the ground? m(e) Find the height from which the ball was thrown. m(f) How long does it take the ball to reach a point 10.0…arrow_forwardA particle initially located at the origin has an acceleration of a⃗ = 3.0ĵm/s2 and an initial velocity of vi = 500îm/s Find (a) the vector position and velocity at any time t and (b) the coordinates and speed of the particle at t = 2.00 s.arrow_forward
- On a spacecraft two engines fire for a time of 389 s. One gives the craft an acceleration in the x direction of ax = 3.41 m/s^2, while the other produces an acceleration in the y direction of ay = 7.34 m/s^2. At the end of the firing period, the craft has velocity components of vx = 1860 m/s and vy = 4290 m/s. Find the (a) magnitude and (b) direction of the initial velocity. Express the direction as an angle with respect to the +x axis.arrow_forwardThe position r→ of a particle moving in an xy plane is given by r→=(3.00t^3−7.00t)i^+(6.00−2.00t^4)j^ with r→ in meters and t in seconds. In unit-vector notation, calculate(a)r→, (b)v→, and (c)a→ for t = 2.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 2.00 s? Give your answer in the range of (-180o; 180o).arrow_forwardA particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t1 = 3.30 s, it is at point (4.40 m, 5.90 m) with velocity (2.70 m/s)ĵ and acceleration in the positive x direction. At time t2 = 12.0 s, it has velocity (–2.70 m/s)î and acceleration in the positive y direction. What are the (a) x and (b) y coordinates of the center of the circular path? Assume at both times that the particle is on the same orbit.arrow_forward
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- A particle moves in the x-y plane with constant acceleration. At time t = 0 s, the position vector for the particle is d = 5.8 m x + 2.1 m y. The acceleration is given by the vector a = 8.3 m/s2 x + 7.8 m/s2 y. The velocity vector at time t = 0 s is v = 5.3 m/s x - 7.5 m/s y. Find the magnitude of the velocity vector at time t = 7.5 s. What is the angle between the velocity vector and the positive x-axis at time t = 7.5 s? What is the magnitude of the position vector at time t = 7.5 s? What is the angle between the position vector and the positive x-axis at time t = 7.5 s?arrow_forwardA particle starts from the origin at t=0, with an initial velocity having an x component of 20m/s and a y component of -15m/s. The particle moves in the xy plane with an x component of acceleration only, given by Ax =4.0 m/s^2 Determine the components of the velocity vector at anytime and the total velocity vector at any time . Calculate the velocity and speed of the particle at t=5.0s.arrow_forwardFrom the origin, a particle starts at t = 0 s with a velocity vecv=7.0hatim/s and moves in the xy plane with a constant acceleration of a=(-9.0hati+3.0hatj)m/s^(2). At the time the particle reaches the maximum x coordinate, what is its (a) velocity and (b) position vector?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning