A helicopter changes its velocity from \[30{m}/{s}\;\] due to north to \[40{m}/{s}\;\] due east in 20 s. What is the magnitude of average acceleration during this time?
Answer
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Hint: The average acceleration equals the change in velocity of a body by the time taken by the body. The change in the value of the velocity of the helicopter can be computed by finding the root of the sum of the squares of the initial and final velocity of the helicopter. As the total time taken by the helicopter is known, so, by dividing the change in velocity of the helicopter by the time taken, we can obtain the value of the average acceleration.
Complete step-by-step solution:
From the given information, we have the data as follows.
A helicopter changes its velocity from \[30{m}/{s}\;\] due to north to \[40{m}/{s}\;\] due east in 20 s.
The average acceleration equals the change in velocity of a body by the time taken by the body. The mathematical representation of the same is,
\[\text{Average acceleration}=\dfrac{\text{Change in velocity}}{\text{Time taken}}\]
Substitute the values in the above equation.
\[\begin{align}
& \text{Average acceleration}=\dfrac{\sqrt{\text{4}{{\text{0}}^{2}}\text{+3}{{\text{0}}^{2}}}}{\text{20}} \\
& \Rightarrow \text{Average acceleration}=\dfrac{\text{50}}{\text{20}} \\
& \therefore \text{Average acceleration}=2.5{m}/{{{s}^{2}}}\; \\
\end{align}\]
\[\therefore \] The magnitude of the average acceleration during 20 s is,\[2.5{m}/{{{s}^{2}}}\;\].
Additional information:
The velocity time graph gives the displacement. When the slope of the velocity time graph will be above the x-axis, then, the displacement is considered to be positive. Similarly, when the slope of the velocity time graph will be below the x-axis, then, the displacement is considered to be negative.
Note: There are different types of graph considering the parameters of the motion with respect to the time, such as, the displacement time graph, velocity time graph and the acceleration time graph. The slope of the displacement time graph gives the velocity, the slope of the velocity time graph gives the acceleration and slope of the acceleration time graph gives the change in the velocity.
Complete step-by-step solution:
From the given information, we have the data as follows.
A helicopter changes its velocity from \[30{m}/{s}\;\] due to north to \[40{m}/{s}\;\] due east in 20 s.
The average acceleration equals the change in velocity of a body by the time taken by the body. The mathematical representation of the same is,
\[\text{Average acceleration}=\dfrac{\text{Change in velocity}}{\text{Time taken}}\]
Substitute the values in the above equation.
\[\begin{align}
& \text{Average acceleration}=\dfrac{\sqrt{\text{4}{{\text{0}}^{2}}\text{+3}{{\text{0}}^{2}}}}{\text{20}} \\
& \Rightarrow \text{Average acceleration}=\dfrac{\text{50}}{\text{20}} \\
& \therefore \text{Average acceleration}=2.5{m}/{{{s}^{2}}}\; \\
\end{align}\]
\[\therefore \] The magnitude of the average acceleration during 20 s is,\[2.5{m}/{{{s}^{2}}}\;\].
Additional information:
The velocity time graph gives the displacement. When the slope of the velocity time graph will be above the x-axis, then, the displacement is considered to be positive. Similarly, when the slope of the velocity time graph will be below the x-axis, then, the displacement is considered to be negative.
Note: There are different types of graph considering the parameters of the motion with respect to the time, such as, the displacement time graph, velocity time graph and the acceleration time graph. The slope of the displacement time graph gives the velocity, the slope of the velocity time graph gives the acceleration and slope of the acceleration time graph gives the change in the velocity.
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