Question

An electron, starting from rest and moving with a constant acceleration, travels 1.0 cm in 2.0 ms. What is the magnitude of this acceleration?

Answer 1

Explanation:

To find the acceleration of the electron we will use a kinematic relation that connects distance covered, acceleration and time $$ s=\frac{1}{2}at^2 $$

Answer 2

Given: Distance = 1 cm

            time = 2ms

           Initial velocity = 0  

To find: Magnitude of acceleration

Solution: According to the law of kinematics,

S = ut + 1/2at²

Where S is the distance travel, u is the initial velocity, a is the acceleration, t is the time taken by the electron to travel the distance.

Since an electron starts from rest, the initial velocity is zero.

1 × 10⁻²m = 0 + 1/2 × a × (2×10⁻³)²

10⁻² = 2 × 10⁻⁶ × a

a= 5 × 10³ ms⁻²

Therefore the magnitude of the acceleration is 5 × 10³ ms⁻²

• Acceleration is the rate of change of velocity with respect to time. It is a vector quantity. It has both magnitude and direction.
• It is the ratio of the force applied to the mass on which force is applied.