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The distance covered by a particle in ti...

The distance covered by a particle in time t is given by `x=a+bt+ct^2+dt^3`, find the dimensions of a,b,c and d.

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To find the dimensions of the constants \(a\), \(b\), \(c\), and \(d\) in the equation \(x = a + bt + ct^2 + dt^3\), we start by recognizing that \(x\) represents distance. ### Step-by-Step Solution: 1. **Identify the Dimension of Distance**: - The left-hand side (LHS) of the equation is \(x\), which represents distance. The dimension of distance is denoted as \(L\) (length). 2. **Analyze the First Term \(a\)**: ...
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Knowledge Check

  • The velocity of a particle (V) at a instant (t) is given by V = at + bt^2 the dimension of b is :

    A
    `L`
    B
    `LT^(-1)`
    C
    `LT^(-2)`
    D
    `LT^(-3)`

  • If the force is given by F = at + bt^(2) with t as time. The dimensions of a and b are

    A
    `MLT^(-4), MLT^(-2)`
    B
    `MLT^(-3) , MLT^(-4)`
    C
    `ML^(2)T^(-3) , ML^(2)T^(-2)`
    D
    `ML^(2)T^(-3) , -ML^(3)T^(-4)`

  • If the force is given by F = at + bt^2 with t as time. The dimensions of a and b are

    A
    `[MLT^(-4)],[MLT^(-2)]`
    B
    `[MLT^(-3)],[MLT^(-4)]`
    C
    `[ML^2T^(-3)],[ML^2T^(-2)]`
    D
    `[ML^2T^(-3)],[ML^3 T^(-4)]`

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    If the force is given by F=at+bt^(2) with t is time. The dimensions of a and b are

    The horizontal and vertical distances covered by a projectile at tiem t are given by x=at and y= bt^(2)+ct , wehre a,b and c are constants. What is the magnitude of the velocity of the projectile 1 second after it is fired?

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