
A particle starts from rest at $ t = 0 $ and moves on a straight line with acceleration as shown graphically. The speed will be minimum at:
(A) $ 1\sec $
(B) $ 2\sec $
(C) $ 3\sec $
(D) $ 4\sec $
Answer
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Hint: We have to observe the graph and understand that this graph is the plot between acceleration and time and we have to consider the relationship between acceleration, velocity and time. We have to use the formulas for acceleration, velocity and time.
Complete answer:
In the above graph, acceleration at time $ t = 0 $ is $ - 7m{s^{ - 2}} $
Let $ a $ be the acceleration, $ t $ be the time and initial velocity be $ u $
Thus velocity $ v $ from $ t = 0 $ to $ t = 2 $ and $ u = 0 $ is given by
$ v = u + at $ …… (kinematic equation)
$ \Rightarrow v = 0 + ( - 7)2 $
$ \Rightarrow v = - 14m{s^{ - 1}} $
Now, from $ t = 2 $ to $ t = 4 $ , $ a = + 7m{s^{ - 2}} $ and initial velocity be $ v = - 14m{s^{ - 1}} $
Hence, the final velocity at $ t = 4 $ is given by
$ v' = v + at $
$ \Rightarrow v' = - 14 + 7(2) $
$ \Rightarrow v' = 0m{s^{ - 1}} $
Thus, the speed at time $ t = 4 $ is minimum.
The correct answer is option D.
Note:
Here, the graph shows the relation between acceleration and time. From this graph we have to find out the minimum velocity at what time with the help of the graph. Here, we have used the relationship given by the first kinematic equation $ v = u + at $ , using this formula we have established the relation between velocity, acceleration and time and put the proper values from the graph and find the required answer. We can also find the maximum velocity but here we have been asked about only minimum velocity. Hence we have calculated the answer with the help of the above procedure. Observe the graph carefully.
Complete answer:
In the above graph, acceleration at time $ t = 0 $ is $ - 7m{s^{ - 2}} $
Let $ a $ be the acceleration, $ t $ be the time and initial velocity be $ u $
Thus velocity $ v $ from $ t = 0 $ to $ t = 2 $ and $ u = 0 $ is given by
$ v = u + at $ …… (kinematic equation)
$ \Rightarrow v = 0 + ( - 7)2 $
$ \Rightarrow v = - 14m{s^{ - 1}} $
Now, from $ t = 2 $ to $ t = 4 $ , $ a = + 7m{s^{ - 2}} $ and initial velocity be $ v = - 14m{s^{ - 1}} $
Hence, the final velocity at $ t = 4 $ is given by
$ v' = v + at $
$ \Rightarrow v' = - 14 + 7(2) $
$ \Rightarrow v' = 0m{s^{ - 1}} $
Thus, the speed at time $ t = 4 $ is minimum.
The correct answer is option D.
Note:
Here, the graph shows the relation between acceleration and time. From this graph we have to find out the minimum velocity at what time with the help of the graph. Here, we have used the relationship given by the first kinematic equation $ v = u + at $ , using this formula we have established the relation between velocity, acceleration and time and put the proper values from the graph and find the required answer. We can also find the maximum velocity but here we have been asked about only minimum velocity. Hence we have calculated the answer with the help of the above procedure. Observe the graph carefully.
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