Solve d = 1/2at^2 for t. Assume all values are positive.
The Correct Answer and Explanation is:
To solve ( d = \frac{1}{2} a t^2 ) for ( t ), we will isolate ( t ) step-by-step.
Step-by-Step Solution
- Start with the given equation:
[
d = \frac{1}{2} a t^2
] - Multiply both sides by 2 to eliminate the fraction:
[
2d = a t^2
] - Divide both sides by ( a ) to isolate ( t^2 ):
[
t^2 = \frac{2d}{a}
] - Take the square root of both sides to solve for ( t ):
[
t = \sqrt{\frac{2d}{a}}
]
Since all values are positive, we don’t need to consider the negative square root, so:
[
t = \sqrt{\frac{2d}{a}}
]
Explanation
In this problem, the equation ( d = \frac{1}{2} a t^2 ) represents the distance ( d ) traveled by an object undergoing constant acceleration ( a ) over time ( t ). This formula is derived from the equations of motion for objects moving under constant acceleration, such as free-falling objects or objects sliding down a frictionless incline. Here, ( d ) is the displacement, ( a ) is the constant acceleration, and ( t ) is the time taken.
To isolate ( t ), we first eliminate the fraction by multiplying both sides by 2. This step is essential for “clearing” the fraction, making the equation easier to manipulate. Next, we divide by ( a ) to remove the acceleration term from the right side, leaving us with ( t^2 ) alone on the right side of the equation. The final step involves taking the square root of both sides, which is how we reverse squaring in algebra.
In physical terms, solving for ( t ) helps us determine how long it takes for an object to travel a specific distance under constant acceleration. This process is common in physics problems, particularly in kinematics, where understanding the relationship between distance, acceleration, and time is crucial. The formula ( t = \sqrt{\frac{2d}{a}} ) allows us to quickly calculate the time when we know the distance and acceleration, which is particularly useful in scenarios like calculating the time it takes for a car to stop or the time a ball takes to reach the ground when dropped from a certain height.