Question

1. Consider a particle moving in a straight line, and assume that its position is defined by the equation x = 6t2 – t3, where t is expressed in seconds and x in meters. a) Obtain the velocity v at any time t by differentiation, dx/dt; b) Obtain the acceleration a at any time t by differentiation dv/dt; c) Draw the x-t, graph v-t, graph a-t until t = 6s; d) What is the average velocity from t = 0 to t = 4

Answer 1

X = 6t² - t³
a) Velocity V at any time t = 12t - 3t²
b) Acceleration A at any time t = 12 - 6t
c) given on the image
d) Average velocity from t =0 to 4 = (9 + 12 +9 +0)/4 = 30/4 = 7.5 m /sec.. ANSWER.
HOPE this is ur required answer.