Question

The acceleration of a particle moving along the x axis may be determined from the expression

a = bt^u + ct^v.

What will be the dimensions of b and c in this case if

u = 10

and

v = 19?

(Use the following as necessary: L for length and T for time.)

Answer 1

Given expression is

a = bt^u + ct^v

dimension of acceleration is

a = m/sec^2 = [L]/[T^2]

dimension of time t is

t = time = sec = [T]

Now in an expression unit of both sides should be same, So

dimension of a = dimension of bt^u

[L]/[T^2] = dimension of b*dimension of t^u

given that u = 10

t^u = t¹⁰ = [T]¹⁰

So

[L]/[T]² = dimension of b*[T]¹⁰

[L]/[T]²*[T]¹⁰ = dimension of b

dimension of b = [L]/[T]^{10 + 2} = [L]/[T]¹² = [L][T]^-12

dimension of a = dimension of ct^v

[L]/[T^2] = dimension of c*dimension of t^v

given that v = 19

t^v = t¹⁹ = [T]¹⁹

So

[L]/[T]² = dimension of c*[T]¹⁹

[L]/[T]²*[T]¹⁹ = dimension of c

dimension of c = [L]/[T]^{19 + 2} = [L]/[T]²¹ = [L][T]^-21

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