Question

A student is walking along the hallway at 1m/s when the bell rings for class. The student reaches their classes their classroom door in 10 seconds ,now moving at 3m/s. How much did they accelerate

Answer 1

The student accelerated at a rate of 0.2 m/s², calculated using the final and initial velocities, and the time taken to change velocity.

Step-by-step explanation:

To calculate the acceleration of the student moving from 1 m/s to 3 m/s in 10 seconds, we can use the formula for acceleration, which is Δv/Δt, where Δv is the change in velocity and Δt is the time taken to change the velocity.

Here, the initial velocity (u) is 1 m/s, the final velocity (v) is 3 m/s, and the time (t) is 10 seconds. The change in velocity (Δv) is v - u = 3 m/s - 1 m/s = 2 m/s. The time (Δt) is 10 seconds. Therefore, the acceleration (Δv/Δt) is 2 m/s / 10 s = 0.2 m/s².

The student therefore accelerated at 0.2 m/s².

Answer 2

final speed of the students when they reached at door of the classroom will be

`v_f = 3 m/s`

initially all are moving at speed

`v_i = 1 m/s`

it took t = 10 s to reach to the classroom

now as per definition of acceleration we will have

`a = (v_f - v_i)/(t)`

now plug in all values in it

`a = (3 - 1)/(10)`

`a = (2)/(10)`

`a = 0.2 m/s^2`

so here student will accelerate at rate of 0.2 m/s^2