Answer
396.9k+ views
Hint: We need to understand the relation between the acceleration of a body and the displacement of the body. The graph of acceleration versus the time graph can give enough information required for the solution that is needed for this problem.
Complete step-by-step solution
We are given a graph of acceleration of a particle versus the time. We know that the slope of certain graphs given the physical parameters related to the quantities on the graph. In our case, the differential of the graph will not give us any required value.
We know that the area under the graph of an acceleration versus time graph can give the velocity of the particle under consideration. We can derive the required relation in terms of time ‘t’ for the velocity using this idea. The integral of the given graph can give the expression for the velocity. So, we can find the equation of the line as –
\[\begin{align}
& y=mx+c \\
& \Rightarrow y=\dfrac{(0-(-2))}{(1-0)}x+-2 \\
& \text{here,} \\
& y=a,x=t \\
& \therefore a=2t-2 \\
\end{align}\]
We can integrate the above equation to get the velocity on the L.H.S as –
\[\begin{align}
& \int\limits_{{}}^{{}}{a}=\int{(2t-2})dt \\
& \Rightarrow v={{t}^{2}}-2t+c \\
& \text{initially,} \\
& v=0 \\
& \therefore v={{t}^{2}}-2t \\
\end{align}\]
This is the required expression for velocity in terms of time ‘t’.
(b) We can find the displacement from the above expression very easily. The displacement is the integral of the velocity with time. So, the displacement of the particle is given as –
\[\begin{align}
& v={{t}^{2}}-2t \\
& \Rightarrow s=\int{v}=\int\limits_{t=2s}^{t=4s}{({{t}^{2}}-2t)}dt \\
& \Rightarrow s=\left[ \dfrac{{{t}^{3}}}{3} \right]_{2}^{4}-[{{t}^{2}}]_{2}^{4} \\
& \therefore s=6.67m \\
\end{align}\]
The displacement of the particle between the two seconds of motion from 2s to 4s is 6.67m. This is the required solution.
Note: The expression for the acceleration of a particle can give the complete details of the motion of a particle as we have seen in this problem. The same is possible for any of the expressions for the displacement of the particle or the velocity of the particle is given.
Complete step-by-step solution
We are given a graph of acceleration of a particle versus the time. We know that the slope of certain graphs given the physical parameters related to the quantities on the graph. In our case, the differential of the graph will not give us any required value.
![seo images](https://www.vedantu.com/question-sets/eca0c854-b189-4e5c-93c3-f3ee48f1b0ea7360063697020266802.png)
We know that the area under the graph of an acceleration versus time graph can give the velocity of the particle under consideration. We can derive the required relation in terms of time ‘t’ for the velocity using this idea. The integral of the given graph can give the expression for the velocity. So, we can find the equation of the line as –
\[\begin{align}
& y=mx+c \\
& \Rightarrow y=\dfrac{(0-(-2))}{(1-0)}x+-2 \\
& \text{here,} \\
& y=a,x=t \\
& \therefore a=2t-2 \\
\end{align}\]
We can integrate the above equation to get the velocity on the L.H.S as –
\[\begin{align}
& \int\limits_{{}}^{{}}{a}=\int{(2t-2})dt \\
& \Rightarrow v={{t}^{2}}-2t+c \\
& \text{initially,} \\
& v=0 \\
& \therefore v={{t}^{2}}-2t \\
\end{align}\]
This is the required expression for velocity in terms of time ‘t’.
(b) We can find the displacement from the above expression very easily. The displacement is the integral of the velocity with time. So, the displacement of the particle is given as –
\[\begin{align}
& v={{t}^{2}}-2t \\
& \Rightarrow s=\int{v}=\int\limits_{t=2s}^{t=4s}{({{t}^{2}}-2t)}dt \\
& \Rightarrow s=\left[ \dfrac{{{t}^{3}}}{3} \right]_{2}^{4}-[{{t}^{2}}]_{2}^{4} \\
& \therefore s=6.67m \\
\end{align}\]
The displacement of the particle between the two seconds of motion from 2s to 4s is 6.67m. This is the required solution.
Note: The expression for the acceleration of a particle can give the complete details of the motion of a particle as we have seen in this problem. The same is possible for any of the expressions for the displacement of the particle or the velocity of the particle is given.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)