Question

The position of an object moving along a line is given by $p(t) = 2{t^3} - 2{t^2} + 1$. What is the speed of the object at $t = 3$ ?

Answer 1

Hint:To solve this type of question, one must know basics of differentiation, we will simply apply differentiation as we know that rate of change of distance is speed so from here, we got the equation of speed with respect to time and then simply substituting the value of time at which we want the speed of the object and hence solving we will get the required solution.

Complete step by step answer:
According to the question the given equation is,
$p(t) = 2{t^3} - 2{t^2} + 1$
And we know that derivative of position is speed,
So,
$
  v(t) = \dfrac{{ds}}{{dt}} = 2{t^3} - 2{t^2} + 1 \\
   \Rightarrow v(t) = 6{t^2} - 4t \\
 $
Now, we have got the equation of speed and according to the question we have to find,
What is the speed of the object at $t = 3$ ?
So, putting the value of time in above equation,
$
  v(t) = 6{t^2} - 4t \\
   \Rightarrow v(3) = 6{(3)^2} - 4(3) \\
   \Rightarrow v(3) = 42m{\text{ }}{s^{ - 1}} \\
 $
Hence, the required speed of the object at $t = 3$ is $42m{\text{ }}{s^{ - 1}}$ .

Note:The rate at which an object moves from one location to another in a certain amount of time is known as speed. It is a scalar quantity since it simply defines the magnitude of a moving item, not its directions. m/s is the SI unit of speed. Note that in the above solution we have used standard units in all cases.