Question

The equation x(t) = −bt² + ct³ gives the position of a particle traveling along the x axis at any time. In this expression, b = 4.00 m/s², c = 4.80 m/s³, and x is in meters when t is entered in seconds. For this particle, determine the following. (Indicate the direction with the sign of your answer as applicable.)

(a) displacement and distance traveled during the time interval t = 0 to t = 3 s

displacement    

distance    

(b) average velocity and average speed during the time interval t = 0 to t = 3 s

average velocity     

average speed     

(c) average acceleration during the time interval t = 0 to t = 3 s

Answer 1

x(t) = −4 t² + 4.8 t³

at t = 0

x(0) = −4 (0)² + 4.8 (0)³ = 0 m

at t = 3

x(3) = −4 (3)² + 4.8 (3)³ = 93.6 m

Displacement = x(3) - x(0) = 93.6 m

x(t) = −4 t² + 4.8 t³

taking derivative relative to "t" both side

dx(t) /dt = - 8 t + 14.4 t²

when velocity is zero

0 = - 8 t + 14.4 t²

t = 0.56 sec

hence particle reverse the direction at t = 0.56 sec

Position at t = 0.56 sec is given as

x(0.56) = −4 (0.56)² + 4.8 (0.56)³ = - 0.411 m

Total distance is given as

distance = 2 X(0.56) + X(3) = 2 (0.411) + 93.6 = 94.42 m

b)

V_avg = average velocity = Displacement/time = 93.6/3 = 31.2 m/s

_vavg = average speed = Distance/time = 94.42/3 = 31.5 m/s

c)

a = change in velocity /time = 31.2 /3 = 10.4 m/s²