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A particle starting from rest attains a ...

A particle starting from rest attains a velocity of `50 ms^(-1)` in 5 s. The acceleration of the particle is

A

`5 m/s^2`

B

`10 m/s^2`

C

`50 m/s^2`

D

`100 m/s^2`

Text Solution

AI Generated Solution

The correct Answer is:
To find the acceleration of the particle, we can use the formula from the first equation of motion: \[ V = U + A \cdot T \] Where: - \( V \) = final velocity - \( U \) = initial velocity - \( A \) = acceleration - \( T \) = time ### Step-by-Step Solution: 1. **Identify the given values:** - Initial velocity, \( U = 0 \, \text{m/s} \) (since the particle starts from rest) - Final velocity, \( V = 50 \, \text{m/s} \) - Time, \( T = 5 \, \text{s} \) 2. **Substitute the known values into the equation:** \[ 50 = 0 + A \cdot 5 \] 3. **Simplify the equation:** \[ 50 = A \cdot 5 \] 4. **Solve for acceleration \( A \):** \[ A = \frac{50}{5} \] \[ A = 10 \, \text{m/s}^2 \] ### Final Answer: The acceleration of the particle is \( 10 \, \text{m/s}^2 \). ---
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Knowledge Check

  • Starting from a stationary position, Sunil moves his car to attain a velocity of 20ms^(-1) in 10s. Find the acceleration of the car.

    A
    `2ms^(1)`
    B
    `2ms^(-2)`
    C
    `2ms^(-1)`
    D
    `2ms^(2)`
  • A particle starts from the origin with a velocity of 10 m s^(-1) and moves with a constant acceleration till the velocity increases to 50 ms^(-1) . At that instant, the acceleration is suddenly reversed. What will be the velocity of the particle, when it returne to the starticng point?

    A
    Zero
    B
    `10ms^(-1)`
    C
    `50 ms^(-1)`
    D
    `70 ms^(-1)`
  • A particle starts moving in +ve x direction with initial velocity of 10 ms^(-1) with a uniform acceleration of magnitude 2 ms^(-2) but directed in -ve x direction. What is the distance traversed by the particle in 12 seconds:

    A
    `-24 m`
    B
    `24 m`
    C
    `70 m`
    D
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