Question

distance covered by an particle in time t is given by x=a+bt+ct²+dt³. find the dimensions of the a,b,c,d.​

Answer 1

Answer:

The dimensions of LHS and RHS should be equal.

Here the dimensions of LHS is the dimensions of length or distance which is [L].

So dimensions of a, bt, ct^2,dt^3 should be [L]

Thus

1) dimension of a=[L]

2) Dimension of bt=[L]

So dimension of b=[L/T]

3) dimension of ct^2=[L]

So dimension of c=[L/T^2]

4) dimension of dt^3=[L]

So dimension of d=[L/T^3]

Answer 2

Answer: a= L

b= LT^-1

c=LT^-2

d=LT^-3

Explanation:

X=a+bt+ct^2+dt^3

From homgenity principle we know that only equal units or dimensions are added and subtracted.

a=x

a=L

bt= x

b = x/t

b = L/T

b = LT^-1

Ct^2 =x

C = x/t^2

C = L/T^2

C = LT^-2

dt^3=x

d = x/t^3

d = L/T^3

d = LT^-3