Mathematics

Mathematics

Problem

The velocity of a body moving along a straight track is given by v(t)=3t2−12t+7  (SI units)v(t)=3t^{2}-12t+7\;(\text{SI units}), where tt is the time in seconds. a. (i) Find an expression for its acceleration as a function of time. b. (ii) Determine the instant(s) when the acceleration becomes zero. c. (iii) State whether the acceleration is positive or negative just before and just after the instant obtained in part (ii).
Intuitive explanation

The key here is understanding that velocity tells you the instantaneous rate of change of position, and acceleration tells you the instantaneous rate of change of velocity. To find when the body changes direction, look for where v(t)=0v(t) = 0—that's when it momentarily stops before reversing. To find when acceleration is zero, take the derivative of velocity and set it equal to zero, which reveals when the body stops speeding up or slowing down.

Intuitive explanation

The key here is understanding that velocity tells you the instantaneous rate of change of position, and acceleration tells you the instantaneous rate of change of velocity. To find when the body changes direction, look for where v(t)=0v(t) = 0—that's when it momentarily stops before reversing. To find when acceleration is zero, take the derivative of velocity and set it equal to zero, which reveals when the body stops speeding up or slowing down.

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