An electron starts at rest. Gravity accelerates the electron in the negative y direction at 9.80 m/s2 while an electric field accelerates it in the positive x direction at 3.80×106 m/s2. Find its velocity 2.45 s after it starts to move.

by the way, the answer is suposed to be in this form: ( , ) m/s. there are supposed to be two answers since there is acceleration in two dimensions.

Velocity , in component form, is

(Vx,Vy) = (at, -gt)
At time t = 2.45 s,
V = (9.30*10^6, -24.0) m/s

To find the velocity of the electron 2.45 s after it starts to move, we can use the kinematic equation that relates velocity, time, and acceleration, which is as follows:

v = u + at

Where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given:
Gravity acceleration (in the negative y direction): -9.80 m/s^2
Electric field acceleration (in the positive x direction): 3.80 × 10^6 m/s^2
Time: 2.45 s

Let's break down the problem and solve it step by step:

1. Calculate the change in velocity due to gravity:
Using the formula v = u + at, where u = 0 m/s (initial velocity) and a = -9.80 m/s^2 (acceleration due to gravity), we can calculate the change in velocity due to gravity over the given time.
v_gravity = 0 + (-9.80) × 2.45

2. Calculate the change in velocity due to the electric field:
Using the formula v = u + at, where u = 0 m/s (initial velocity) and a = 3.80 × 10^6 m/s^2 (acceleration due to the electric field), we can calculate the change in velocity due to the electric field over the given time.
v_electricField = 0 + (3.80 × 10^6) × 2.45

3. Calculate the resultant velocity:
The resultant velocity is the vector sum of the velocities due to gravity and the electric field, taking into account their respective directions.
The resultant velocity can be found using the Pythagorean theorem.
resultant_velocity = √(v_gravity^2 + v_electricField^2)

Now, let's calculate the values step by step:

1. v_gravity = 0 + (-9.80) × 2.45
v_gravity = -23.941 m/s (rounded to three decimal places)

2. v_electricField = 0 + (3.80 × 10^6) × 2.45
v_electricField = 9.29 × 10^6 m/s (rounded to three significant figures)

3. resultant_velocity = √((-23.941)^2 + (9.29 × 10^6)^2)
resultant_velocity = √(574.38 + 8.63 × 10^13)
resultant_velocity = √(8.63 × 10^13)
resultant_velocity = 9.28 × 10^6 m/s (rounded to three significant figures)

Therefore, the velocity of the electron 2.45 s after it starts to move is approximately 9.28 × 10^6 m/s in the positive x direction.