A brick is dropped freely from the roof of a 20m tall building. Air resistance may be ignored. What is the magnitude of the brick's velocity just before it reaches the ground?
a (acceleration g) = 9.8 m/s²
d (distance ) = 20 m
Vi (initial velocity) = 0 m/s
Vf (final velocity) = ? unknown
To determine tha magnitude of brick's velocity, we will first determine the final velocity (Vf) and we can use the formula Vf² = Vi² + 2ad
Since Vi = 0 m/s we will just used Vf = √2ad
Vf = √2ad
= √2 × 9.8 m/s² × 20 m
= √ 392 m²/s²
= 19.799 m/s
From 0 m/s to 19.799 m/s
The magnitude of brick's velocity is 19.799 m/s
50kg = 6m, 1m = ? 50kg ×1m ÷ 6 = 8.3kg
During the formation of the universe some 14 billion years ago in the so- called 'BigBang' , only the lightest elements were formed.
The distance of the car is 6 meters.
In our case, a car is travelling with an initial velocity then eventually slows down when the driver apply brakes. This type of solving problem is about Motions in a Straight Line.
Here, we are going to solve for the distance traveled by the car. To solve this, we are going to use the formula
and are the final and initial velocities respectively, unit is in m/s
is the acceleration, unit is in m/
is the horizontal distance traveled by the car, unit is in meter (m)
For the given information
For the first scenario of the car, here are the following given:
= 7 m/s
x = 1.5 m
For the second scenario, the same car have this given:
= 14 m/s
x = ?
Solving the problem
1. First, we will solve for the acceleration of the car of the first scenario using the formula above
Substitute the given information, we have
Transpose to the left side of the equation, we have:
Simplify, then divide both side of the equation by , we have:
a = - 16.33 m/
The negative sign for the acceleration means that the car is slowing down (or the term is deceleration) because the brakes were applied.
2. Now, we are ready to solve for the distance traveled by the car on the second scenario using the same formula.
Substitute the given information for the second scenario, we have:
Simplify the equation
Transpose to the left side of the equation and solve for , we have
Divide both sides of the equation by 32.67, we have:
x = 6 meters
Therefore, the distance traveled by the car is 6 meters for the second scenario.
To learn more about motions in a straight line, just click on the following links:Additional example
Formula for acceleration