Question

A particle is in motion and is accelerating. The functional form of the velocity is v(t) = 20t − 5t2 m/s . f. Interpret the results of (c) in terms of the directions of the acceleration and velocity vectors. (c. Find Position of the particle at 1, 2, 3, and 5 s. S(1)=10(1)2- (5(1)3)/3= 8.33m S(2)=10(2)2- (5(2)3)/3= 26.67m S(3)=10(3)2- (5(3)3)/3= 45m S(5)=10(5)2- (5(5)3)/3= 41.67m

Name: a particle is in motion and is accelerating the functional form of the velocity is vt 20t 5t2 ms f interpret the results of c in terms of the directions of the acceleration and velocity vect 69002
Uploaded: 2023-01-02T13:51:42-08:00
Duration: 3 min 33 s
Channel: Sri K
Description: a particle is in motion and is accelerating the functional form of the velocity is vt 20t 5t2 ms f interpret the results of c in terms of the directions of the acceleration and velocity vect 69002

          A particle is in motion and is accelerating. The functional form
of the velocity is v(t) = 20t − 5t2 m/s .
f. Interpret the results of (c) in terms of the directions of
the acceleration and velocity vectors.
(c. Find Position of the particle at 1, 2, 3, and 5 s.
S(1)=10(1)2- (5(1)3)/3= 8.33m
S(2)=10(2)2- (5(2)3)/3= 26.67m
S(3)=10(3)2- (5(3)3)/3= 45m
S(5)=10(5)2- (5(5)3)/3= 41.67m

Added by David N.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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A particle is in motion and is accelerating. The functional form of the velocity is v(t) = 20t − 5t2 m/s . f. Interpret the results of (c) in terms of the directions of the acceleration and velocity vectors. (c. Find Position of the particle at 1, 2, 3, and 5 s. S(1)=10(1)2- (5(1)3)/3= 8.33m S(2)=10(2)2- (5(2)3)/3= 26.67m S(3)=10(3)2- (5(3)3)/3= 45m S(5)=10(5)2- (5(5)3)/3= 41.67m