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Initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. two seconds later, its velocity is 6.11 m/s at an angle of 54.2° below the horizontal. what was the particle's average acceleration during these 2.00 seconds in the x-direction (enter first) and the y-direction?

The x-coordinate of a particle in curvilinear motion is given by x = 3.1t3 - 4.9t where x is in feet and t is in seconds. the y-component of acceleration in feet per second squared is given by ay = 2.3t. if the particle has y-components y = 0 and vy = 5.0 ft/sec when t = 0, find the magnitudes of the velocity v and acceleration a when t = 1.8 sec. sketch the path for the first 1.8 seconds of motion, and show the velocity and acceleration vectors for t = 1.8 sec. answers: v = ft/sec a = ft/sec2

Aparticle initially moving east with a speed of 20.0 m/s, experiences an acceleration of 3.95 m/s, north for a time of 8.00 s. what was the speed of the particle after this acceleration, in units of m/s? give the answer as a positive number.