
A body travels $ 200cm $ in the first two seconds and $ 220cm $ in the next four seconds. What will be the velocity at the end of seventh second from the start?
Answer
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Hint
This problem can be solved by using equations of motion. From the given conditions, calculate the values of initial velocity and acceleration of the body and then by using these values, the velocity at the end of seventh second can be found.
Formulas Used:
$ S = ut + \dfrac{1}{2}a{t^2} $
$ v = u + at $
Complete step by step answer
Case 1:
Distance travelled by the body = $ 200cm $
Time taken = $ 2\sec $
From equations of motion, we have the formula
$ S = ut + \dfrac{1}{2}a{t^2} $
By substituting the given values in the above formula, we get
$ \Rightarrow 200 = u(2) + \dfrac{1}{2}a{(2)^2} $
$ \Rightarrow 200 = 2u + 2a $
On further simplification, we get
$ \Rightarrow u + a = 100 $ ….. (1)
Case 2:
In the next four seconds the body travels a distance of $ 220cm $ . The total distance travelled and total time taken from the start of the motion will be
Total distance travelled by the body = $ 200 + 220 = 420cm $
Total time taken = $ 2 + 4 = 6\sec $
From equations of motion, we have the formula
$ S = ut + \dfrac{1}{2}a{t^2} $
By substituting the given values in the above formula, we get
$ \Rightarrow 420 = u(6) + \dfrac{1}{2}a{(6)^2} $
$ \Rightarrow 420 = 6u + 18a $
On further simplification, we get
$ \Rightarrow u + 3a = 70 $ ….. $ (2) $
By simplifying equations $ (1) $ and $ (2) $ , we get
$ \Rightarrow u = 115cm/s $
$ \Rightarrow a = - 15cm/{s^2} $
Case 3:
Total time taken = $ 7\sec $
Now to find the velocity of the body at the end of the $ {7^{th}} $ second, we use the formula
$ v = u + at $
By substituting the values of initial velocity and acceleration in the above formula
$ \Rightarrow v = 115 + ( - 15)(7) $
$ \Rightarrow v = 115 - 105 $
$ \Rightarrow v = 10m/s $
The velocity of the body at the end of seventh second from the start is $ 10m/s $
Note
While substituting the values of distance travelled and time taken in the formula in case 2, the total distance and total time taken from the start of the motion should be considered, so that the initial velocity will be the same in both the conditions.
This problem can be solved by using equations of motion. From the given conditions, calculate the values of initial velocity and acceleration of the body and then by using these values, the velocity at the end of seventh second can be found.
Formulas Used:
$ S = ut + \dfrac{1}{2}a{t^2} $
$ v = u + at $
Complete step by step answer
Case 1:
Distance travelled by the body = $ 200cm $
Time taken = $ 2\sec $
From equations of motion, we have the formula
$ S = ut + \dfrac{1}{2}a{t^2} $
By substituting the given values in the above formula, we get
$ \Rightarrow 200 = u(2) + \dfrac{1}{2}a{(2)^2} $
$ \Rightarrow 200 = 2u + 2a $
On further simplification, we get
$ \Rightarrow u + a = 100 $ ….. (1)
Case 2:
In the next four seconds the body travels a distance of $ 220cm $ . The total distance travelled and total time taken from the start of the motion will be
Total distance travelled by the body = $ 200 + 220 = 420cm $
Total time taken = $ 2 + 4 = 6\sec $
From equations of motion, we have the formula
$ S = ut + \dfrac{1}{2}a{t^2} $
By substituting the given values in the above formula, we get
$ \Rightarrow 420 = u(6) + \dfrac{1}{2}a{(6)^2} $
$ \Rightarrow 420 = 6u + 18a $
On further simplification, we get
$ \Rightarrow u + 3a = 70 $ ….. $ (2) $
By simplifying equations $ (1) $ and $ (2) $ , we get
$ \Rightarrow u = 115cm/s $
$ \Rightarrow a = - 15cm/{s^2} $
Case 3:
Total time taken = $ 7\sec $
Now to find the velocity of the body at the end of the $ {7^{th}} $ second, we use the formula
$ v = u + at $
By substituting the values of initial velocity and acceleration in the above formula
$ \Rightarrow v = 115 + ( - 15)(7) $
$ \Rightarrow v = 115 - 105 $
$ \Rightarrow v = 10m/s $
The velocity of the body at the end of seventh second from the start is $ 10m/s $
Note
While substituting the values of distance travelled and time taken in the formula in case 2, the total distance and total time taken from the start of the motion should be considered, so that the initial velocity will be the same in both the conditions.
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