Question

Which of the following quantities have the dimensions of a speed?

(1)at (2)at^2 (3) (2ax)^(1/2) (4)((2x)/a)^(1/2)
Dimension of speed =
Other dimension =

Answer 1

The quantities at and (2ax)^(1/2) have dimensions of speed because both, when calculated dimensionally, result in L/T or LT^-1, which is the dimensional representation of speed.

Step-by-step explanation:

The question is asking which of the given quantities have dimensions that could represent speed. The dimension of speed is given by L/T or LT-1, which means length divided by time. Speed itself is defined as the distance traveled over time, or ds/dt. Given that the dimensions provided for s (displacement) are [s] = L, and the dimensions of t (time) are [t] = T, any quantity that has dimensions of L multiplied or divided by T to the power of 1 is dimensionally equivalent to speed.

Using this information, we can analyze the given quantities:

1. at: Given [a] = LT-2 and [t] = T, at does have dimensions of L, but since [at] = LT-2T = LT-1, it represents speed.
2. at2: Given [a] = LT-2 and [t2] = T2, at2 has dimensions L, but since [at2] = LT-2T2 = LT, it does not represent speed.
3. (2ax)1/2: Given [a] = LT-2 and [x] = L, (2ax) has dimensions L2T-2, so [(2ax)1/2] = (L2T-2)1/2 = LT-1, which represents speed.
4. ((2x)/a)1/2: Given [x] = L and [a] = LT-2, (2x/a) has dimensions T2, so [((2x)/a)1/2] = (T2)1/2 = T, which does not represent speed.

Therefore, the quantities that have dimensions of a speed are at and (2ax)1/2.

Answer 2

(1) and (3)

Step-by-step explanation:

Speed has dimensions of:

m/s (meters per second)

While:

a (acceleration) has dimensions of
`m/s^2`

t (time) has dimensions of
`s`

Let's analyze each option:

(1)at

`(m/s^2) \cdot (s) = m/s` --> this has dimensions of speed

(2)at^2

`(m/s^2) \cdot (s)^2 = m` --> this has dimensions of distance

(3) (2ax)^(1/2)

`√((m/s^2)\cdot (m))=m/s` --> this has dimensions of speed

(4) ((2x)/a)^(1/2)

`\sqrt{(m)/(m/s^2)}=s` --> this has dimensions of time

So choices (1) and (3) are correct.