Question

The equation *x*(*t*) = −*bt*^{2} +
*ct*^{3} gives the position of a particle traveling
along the *x* axis at any time. In this expression,
*b* = 4.00 m/s^{2}, *c* = 4.80
m/s^{3}, 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 t^{2} + 4.8 t^{3}

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 t^{2} + 4.8 t^{3}

taking derivative relative to "t" both side

dx(t) /dt = - 8 t + 14.4 t^{2}

when velocity is zero

0 = - 8 t + 14.4 t^{2}

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/s^{2}

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