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A particle is moving along X-axis accord...

A particle is moving along X-axis according to the following equation: `x=u(t-4)+a(t-3)^(2)`
All terms in above equation are measured in MKS system

A

Acceleration of the particle is a.

B

Acceleration of the particle is 2a.

C

Velocity of particle at t=3 s is u.

D

At t=0 particle is at the origin.

Text Solution

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The correct Answer is:
To solve the problem step by step, we will analyze the given equation of motion and derive the required quantities: acceleration and velocity. ### Given Equation: The equation of motion for the particle is given as: \[ x = u(t - 4) + a(t - 3)^2 \] ### Step 1: Find the Velocity Velocity \( v \) is defined as the rate of change of displacement with respect to time. We can find the velocity by differentiating the displacement equation with respect to time \( t \). #### Differentiate the equation: \[ v = \frac{dx}{dt} \] Using the product rule and chain rule: 1. Differentiate \( u(t - 4) \): \[ \frac{d}{dt}[u(t - 4)] = u \] 2. Differentiate \( a(t - 3)^2 \): \[ \frac{d}{dt}[a(t - 3)^2] = a \cdot 2(t - 3) = 2a(t - 3) \] Combining these results, we have: \[ v = u + 2a(t - 3) \] ### Step 2: Find the Velocity at \( t = 3 \) seconds Now, we need to find the velocity at \( t = 3 \) seconds: \[ v(t = 3) = u + 2a(3 - 3) \] \[ v(t = 3) = u + 2a(0) \] \[ v(t = 3) = u \] ### Step 3: Find the Acceleration Acceleration \( a \) is defined as the rate of change of velocity with respect to time. We can find acceleration by differentiating the velocity equation with respect to time \( t \). #### Differentiate the velocity equation: \[ a = \frac{dv}{dt} \] From our previous velocity equation: \[ v = u + 2a(t - 3) \] Differentiating: 1. The term \( u \) is constant, so its derivative is \( 0 \). 2. Differentiate \( 2a(t - 3) \): \[ \frac{d}{dt}[2a(t - 3)] = 2a \] Thus, we have: \[ a = 2a \] ### Conclusion - The velocity of the particle at \( t = 3 \) seconds is \( u \). - The acceleration of the particle is \( 2a \). ### Final Answers: - **Velocity at \( t = 3 \) seconds:** \( u \) - **Acceleration:** \( 2a \)
To solve the problem step by step, we will analyze the given equation of motion and derive the required quantities: acceleration and velocity. ### Given Equation: The equation of motion for the particle is given as: \[ x = u(t - 4) + a(t - 3)^2 \] ### Step 1: Find the Velocity Velocity \( v \) is defined as the rate of change of displacement with respect to time. We can find the velocity by differentiating the displacement equation with respect to time \( t \). ...
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Knowledge Check

  • A particle moves along x-axis according to the law x=(t^(3)-3t^(2)-9t+5)m . Then :-

    A
    In the interval `3 lt t lt 5`, the particle is moving in +x direction
    B
    The particle reverses its direction of motion twice in entire motion if it starts at t=0
    C
    The average acceleration from `1 le t le 2` second is `6 m//s^(2)`
    D
    In the interval `5 le t le 6` seconds, the distance travelled is equal to the displacement.

  • A particle moves on the X- axis according to the equation x = x_(0) sin^(2)omegat . The motion is simple harmonic

    A
    with amplitude `x_(0)//2`
    B
    with amplitude `2x_(0)`
    C
    with the period `(2x)/(omega)`
    D
    with the period `(pi)/(omega)`

  • A particle moves along the X- axis according to the equation x = 10 sin^(3)(pit) . The amplitudes and frequency of component SHMs are.

    A
    amplitude `30//4, 10//4 :` frequencies `3//2, 1//2`
    B
    amplitude `30//4, 10//4 :` frequencies `1//2, 3//2`
    C
    amplitude `10, 10 ,` frequencies `1//2, 1//2`
    D
    amplitude `30//4, 10 :` frequencies `3//2, 2`