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At t = 0, a particle moving in the xy plane with constant acceleration has a velocity of
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Chapter 3 Solutions
Principles of Physics: A Calculus-Based Text
- A particle P moves along the line x+y=2 in the z plane with a uniform speed 3√2 ft /sec from the point z = −5 + 7i to the point z = 10−8i. If w = 2z^(2)−3 and P′ is the image, in the plane w, of P, find the magnitude of a) the velocity and b) of the acceleration of P′ After 3 secondsarrow_forwardA particle undergoes three consecutive displacements d1=(1.5i+3.0j-1.2km) cm, d2=(2.3i-1.4j-3.6k) cm and d3=(-1.3i+1.5j) cm. find the component and its magnitude.arrow_forwardA particle P travels with constant speed on a circle of radius r =3.00 m and completes one revolution in 20.0 s. The particle passes through O at time t = 0. State the following vectors in magnitudeangle notation (angle relative to the positive direction of x).With respect to O, find the particle’s position vector at the times t of (a) 5.00 s, (b) 7.50 s, and (c) 10.0 s. (d) For the 5.00 s interval from the end of the fifth second to the end of the tenth second, find the particle’s displacement. For that interval, find (e) its average velocity and its velocity at the (f) beginning and (g) end. Next, find the acceleration at the (h) beginning and (i) end of that interval.arrow_forward
- A particle is launched from the origin in R^3 with initial velocity v(0) = < 0, 8, 10 > and undergoes constant acceleration a = < 2, - 1, - 5 > due to the combined forces of gravity and wind. Where and at what time does the particle hit the "ground" (e.g. the plane z = 0)?arrow_forwardHere are three vectors in meters:d→1=-3.70î+7.00ĵ+1.40k̂d→2=-2.00î-4.00ĵ+2.00k̂d→3=2.00î+3.00ĵ+1.00k̂What results from (a) d→1⋅(d→2+d→3), (b) d→1⋅(d→2×d→3), and d→1×(d→2+d→3) ((c), (d) and (e) for î, ĵ and k̂ components respectively)?arrow_forwardAt t = 0, a particle leaves the origin with a velocity of 9.0 m/s in the positive y direction and moves in the xyplane with a constant acceleration of (2.0i −4.0j) m/s2. At the instant the x coordinate of the particle is 15 m, what is the speed of the particle (in m/s)?arrow_forward
- 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_forwardThe coordinates of a particle moving in the XY plane are given as function of time by: X = 5m + ( 7 m/s^4) t ^4 Y = ( 10 m/s)t + (3 m/s^5)t ^5 Find the position vector of the body at t = 3.5 seconds and instantaneous acceleration of the body at t = 3.5 secarrow_forwardA particle moves in the xy plane in a circle centered at the origin. At a certain instant the velocity and acceleration of the particle are 6.0i m / s and (3.0i + 4.0j) m / s2 respectively. What are the x and y coordinates of the particle at this time?arrow_forward
- For the following three vectors, what is 3⋅C→⋅(2A→×B→) ?A→=4.00î+2.00ĵ-3.00k̂B→=-3.00î+4.00ĵ+3.00k̂C→=6.00î-8.00ĵarrow_forwarda heavy piece of machinery is raised by sliding it a distance d = 12.5 m along a plank oriented at angle u = 20.0° to the horizontal. How far is it moved (a) vertically and (b) horizontally?arrow_forwardA fighter plane moving 211.0 m/s horizontally fires a projectile with speed 47 m/s in a forward direction 30.0° below the horizontal. What is the speed of the projectile with respect to a stationary observer on the ground? a) 240 m/s b) 283 m/s c) 253 m/s d) 278 m/sarrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning