The equation x ( t ) = bt 2 + ct 3 gives the position of a particle traveling al

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

displacement    

distance    

Explanation / Answer

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)

Vavg = 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

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