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QUESTION FIVE A. Define the following terms or words (i) Displacement (ii) Velocity (iii) Speed (iv) acceleration (v) time B. prove if the following equation is dimensionally correct (i). V=U +at

Physics

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Here are the answers to Question Five:

A. Define the following terms or words

Part i

Displacement is a vector quantity that represents the shortest distance and direction from an object's initial position to its final position.

Part ii

Velocity is a vector quantity that describes the rate at which an object changes its position, including both its speed and direction.

Part iii

Speed is a scalar quantity that describes how fast an object is moving, defined as the magnitude of its velocity or the rate at which it covers distance.

Part iv

Acceleration is a vector quantity that describes the rate at which an object's velocity changes over time, including changes in speed or direction.

Part v· 5 steps

Time is a fundamental scalar quantity that measures the duration of events and the interval between them, representing the progression from past to present to future.

B. Prove if the following equation is dimensionally correct

The equation is V=U+atV = U + at.
To prove dimensional correctness, we must show that the dimensions of each term in the equation are the same.

  1. Identify the dimensions of each variable.

    • Dimension of velocity (VV or UU): [L][T]−1[L][T]^{-1} (Length per unit Time)
    • Dimension of acceleration (aa): [L][T]−2[L][T]^{-2} (Length per unit Time squared)
    • Dimension of time (tt): [T][T] (Time)
  2. Determine the dimension of the left-hand side (LHS).

    The LHS is VV.

    [LHS]=[V]=[L][T]−1[LHS] = [V] = [L][T]^{-1}
  3. Determine the dimension of the first term on the right-hand side (RHS).

    The first term on the RHS is UU.

    [U]=[L][T]−1[U] = [L][T]^{-1}
  4. Determine the dimension of the second term on the right-hand side (RHS).

    The second term on the RHS is atat.

    [at]=[a]×[t][at] = [a] \times [t]

    Substitute the dimensions of aa and tt:

    [at]=([L][T]−2)×([T])[at] = ([L][T]^{-2}) \times ([T]) [at]=[L][T]−2+1[at] = [L][T]^{-2+1} [at]=[L][T]−1[at] = [L][T]^{-1}
  5. Compare the dimensions of all terms.

    For the equation to be dimensionally correct, the dimensions of all terms added or subtracted must be identical, and equal to the dimension of the term on the other side of the equation.

    [LHS]=[V]=[L][T]−1[LHS] = [V] = [L][T]^{-1} [U]=[L][T]−1[U] = [L][T]^{-1} [at]=[L][T]−1[at] = [L][T]^{-1}

    Since all terms have the same dimension, [L][T]−1[L][T]^{-1}, the equation is dimensionally correct.

    The equation V=U+atV = U + at is dimensionally correct.

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