A particle moves along a straight line with equation of moti

Solution

The key to any equation of motion problem is to realize that you must take the derivative of the \'position\' function to get the \'velocity\' function f(t) = (t^-1) - t f\'(t) = d/dt(t^-1) - t) = d/dt(t^-1) - d/dt(t) f\'(t) = -1(t^-2) -1 = -t^-2 - 1 f\'(t) = v(t), which we can rewrite as -1/t^2 - 1 (if you like) So v(6) = -1/(5^2) - 1 = -1/25- 1 = -26/25 It should be pointed out (to your teacher perhaps for extra credit) that the dimension of this problem doesn\'t make much sense. In a polynomial description of distance, f(t), you have a function with no constants to carry your dimensions and thus the equation is unbalanced. What this means is: s = distance = f(t). All of the terms in the equation should have units of distance. 1/t, your first term, has units of 1/time. -t, the second term, has dimensions of time. Really, this function needs to be written as f(t) = A(t^-1) - Bt Where A has units of time * distance and B has units of distance/time. This all being said just so you realize that units are important in these types of problems. Your velocity should be -26/25 m/s because you were given meters and seconds in the problem. The minus sign is important because it indicates direction, making it a velocity. The \'speed\' is just 26/25 m/s