Due to an acceleration of 2 m/s 2, the velocity of a body increases from 20 m/s to 30 m/s in a certain period. Find the displacement (in m) of the body in that period.
125
Calculating Displacement with Constant Acceleration
This problem asks us to find the displacement of a body given its initial velocity, final velocity, and constant acceleration. We can use the equations of kinematics for constant acceleration to solve this.
Understanding the Given Information
We are provided with the following information:
- Initial velocity (\(u\)) = 20 m/s
- Final velocity (\(v\)) = 30 m/s
- Acceleration (\(a\)) = 2 m/s\(^2\)
We need to find the displacement (\(s\)) of the body during this period.
Choosing the Right Kinematic Equation
We have initial velocity (\(u\)), final velocity (\(v\)), and acceleration (\(a\)), and we need to find displacement (\(s\)). The kinematic equation that relates these four quantities is:
\[v^2 = u^2 + 2as\]
This equation is suitable because it does not require knowing the time period (\(t\)), which is not given in the problem.
Step-by-Step Calculation of Displacement
We need to rearrange the equation \(v^2 = u^2 + 2as\) to solve for \(s\). Subtract \(u^2\) from both sides:
\[v^2 - u^2 = 2as\]
Now, divide both sides by \(2a\) to isolate \(s\):
\[s = \frac{v^2 - u^2}{2a}\]
Now, substitute the given values into this equation:
- \(v = 30 \, \text{m/s}\)
- \(u = 20 \, \text{m/s}\)
- \(a = 2 \, \text{m/s}^2\)
So, the equation becomes:
\[s = \frac{(30 \, \text{m/s})^2 - (20 \, \text{m/s})^2}{2 \times 2 \, \text{m/s}^2}\]
Calculate the squares:
- \( (30)^2 = 900 \)
- \( (20)^2 = 400 \)
Substitute these values back:
\[s = \frac{900 - 400}{4}\]
Calculate the difference in the numerator:
\[s = \frac{500}{4}\]
Finally, perform the division:
\[s = 125 \, \text{m}\]
Thus, the displacement of the body in that period is 125 meters.
Final Answer Summary
Given an initial velocity of 20 m/s, a final velocity of 30 m/s, and a constant acceleration of 2 m/s\(^2\), the displacement calculated using the kinematic equation \(v^2 = u^2 + 2as\) is 125 meters.
| Quantity | Symbol | Value |
|---|---|---|
| Initial Velocity | \(u\) | 20 m/s |
| Final Velocity | \(v\) | 30 m/s |
| Acceleration | \(a\) | 2 m/s\(^2\) |
| Displacement | \(s\) | ? |
Revision Table: Kinematic Equations
Here are the primary kinematic equations used for motion with constant acceleration:
| Equation | Variables Included |
|---|---|
| \(v = u + at\) | Final velocity, Initial velocity, Acceleration, Time |
| \(s = ut + \frac{1}{2}at^2\) | Displacement, Initial velocity, Acceleration, Time |
| \(v^2 = u^2 + 2as\) | Final velocity, Initial velocity, Acceleration, Displacement |
| \(s = \left(\frac{u+v}{2}\right)t\) | Displacement, Initial velocity, Final velocity, Time |
Additional Information: Constant Acceleration
Constant acceleration means that the velocity of the body changes by the same amount in every equal interval of time. This is a fundamental concept in kinematics.
- When acceleration is in the same direction as velocity, the body speeds up.
- When acceleration is in the opposite direction to velocity, the body slows down (deceleration).
- The kinematic equations listed above are only applicable when the acceleration is constant. If acceleration varies, calculus must be used.
Understanding the relationship between displacement, velocity, acceleration, and time under constant acceleration is crucial for solving many physics problems.
Important Questions from Motion
An object is covering distance in direct proportion to the square of time elapsed. What conclusion can be drawn about the motion of the object?
If the distance time graph of the motion of an object is a straight line but not parallel to the time axis, then it may be concluded that the object is moving with a:
Which of the following changes when a body performs uniform circular motion?
Vehicles have treaded tires so that it_______.
Which of the following is correct with regard to friction?
