The displacement of a particle is given by S = a+bt+ct^2. Deduce the dimensions of the constants,a,b,and c appearing in the question.
Haven't I seen this before?
a is m
b is m/s
c is m/s^2
yes
To deduce the dimensions of the constants a, b, and c appearing in the question, we need to analyze the dimensions of the displacement equation S = a + bt + ct^2.
The equation for displacement has the dimensions of length, so let's start with the dimensions of each term in the equation.
The first term, 'a', represents a constant. Since displacement has dimensions of length, 'a' must also have dimensions of length.
The second term, 'bt', represents the product of a constant 'b' and the variable 't', which represents time. Time has dimensions of seconds (s), so we know that 'bt' has dimensions of length. To find the dimensions of 'b', we need to divide 'bt' by 't', which gives us b = [bt]/[t] = [L]/[T] = L/T, where L represents length and T represents time.
The third term, 'ct^2', represents the product of a constant 'c' and 't^2'. The dimensions of 't^2' are (s)^2, so to maintain the dimensions of length for the displacement equation, 'c' must have dimensions of length divided by time squared, i.e., c = [ct^2]/[t^2] = [L]/[T^2] = L/T^2.
Hence, we can summarize the dimensions of the constants as:
- The constant 'a' has dimensions of length (L).
- The constant 'b' has dimensions of length divided by time (L/T).
- The constant 'c' has dimensions of length divided by time squared (L/T^2).