10. A small toy airplane is flying in the xy- plane parallel to the ground. During the time interval t, 0s and t2 1s, its velocity as a function of time is given by i=1.2 s2 - 2 -- At what values of t is the velocity of the plane perpendicular to its acceleration?
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Q: 10. A small toy airplane is flying in the xy – plane parallel to the ground. During the time…
A: Given,
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- The position of a particle in space at time tis: r(t) = (sec(t)) * i + (tan t) * j + 4/3 tk. Write the particle's velocity at time t = (pi / 6) as the product of its speed and direction.The equation r(t) = ( sin t)i + ( cos t)j + (t) k is the position of a particle in space at time t. Find the particle's velocity and acceleration vectors. π Then write the particle's velocity at t= as a product of its speed and direction. The velocity vector is v(t) = (i+j+ k.The position of an object as a function of time is given by r = (3.2t + 1.5t2)i + (1.7t -2.0t2)jm, where t is time in seconds. What is the magnitude of the acceleration, and what is the direction of the acceleration (given in terms of theta, clockwise from the x axis).
- 1. The dog runs across the backyard lot on which a set of coordinates axes has, strangely observed, been drawn. The coordinates of the dog’s position as functions of time t are given by X = -0.35?2 + 7.5t +30 and Y = 0.25?2 – 9.5t + 26. With t in seconds and x and y in meters.(a) At t =15.0s, what is the dogs position vector ? in unit – vector notation and as magnitude and an angle. (b) The velocity ? at time t =15.0s in unit – vector notation and as magnitude and an angle. (c) Find the acceleration ? at time t = 15.0s in unit – vector notation and as magnitude and an angle.A jeep travels a distance d=22.1m in the positive x direction in a time t1=20.2s, at which point the jeep brakes, coming to rest in t2=7.38s. 1. What was the jeep's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? 2. Using the result from question 1, what was the jeep's horizontal component of acceleration, in meters per squared second, during the braking period?5. A particle moves counterclockwise on a circular path of 400 ft radius. It starts from a fixed point which is horizontally to the right of the center of the path and moves so that s=1012 + 20t where s is the arc distance in feet and t is the time in sec. Compute the x and y components of acceleration at the end of 3 sec.
- A particle moves through 3-space in such a way that its acceleration is a(t)= 8sin 2ri +8cos 2tj+e'k . where t is time measured in seconds. The initial velocity of the particle is v, = 2i – 3j+k. At time t = t , find i) the velocity vector of the particle. ii) the scalar tangential and normal components of acceleration.The range of the projectile motion R in meters is given by the formula vo2 sin 20 where vo m/s is the initial velocity, g m/s² is the R = acceleration due to gravity and 0 is the radian measure of the angle makes with the horizontal at which the projectile is launched. What is the value of 0 that makes the range maximum? 3 4 6. the height of the right circular cylinder of greatest volume that can be inscribed inside the right circular cone with radius 4 in and height 12 in.A particle is moving in three dimensions and its position vector is given by; r(t) = (4t² + 1.7t) î + (1.5t − 2.1)ĵ + (2.7t³ + 2t) k where r is in meters and t is in seconds. Determine the magnitude of the instantaneous velocity at t = 3s. Express your answer in units of m/s using one decimal place. Answer:
- An object is moving with an initial velocity of 5.8 i m/s and a final velocity 13.8 i m/s. The time taken for acceleration is 2.2 seconds. Calculate the acceleration in m/s2 to 2 sf. In your answer, you do not enter the i unit vector, but you do need to enter negative signs, if appropriate. e.g. if your answer is a = -5.4 i m/s,2 you would enter -5.42 = 2500 cos 20 13. For a short time, the jet plane moves along a path in the shape of a lemniscate, r² = (2500cos 26) km². At the instant 0=30°, the radar tracking device is rotating at O =5(103) rad/s with ö=2(103) rad/s². Determine the components of the velocity and acceleration of the plane at this instant, a) in Cartesian Coordinates, b) in Normal and Tangential Coordinates, c) in Polar Coordinates. d) Also calculate the radius of curvature at the same instant.In using polar coordinates to describe plane polar motion the velocity is expressed as: O v = rð e, + †0e, O v = r0e, + rðe, O v = rð e, + re, O v = r e, + r0e,