Answer
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- Hint:- We had to only compare the given distance with the formula to find the distance travelled in nth second i.e. \[d = u + a\left( {n - \dfrac{1}{2}} \right)\]. From this we will get the value of a (acceleration) and u (initial velocity).
Formula used :- Distance travelled in n seconds = \[u + a\left( {n - \dfrac{1}{2}} \right)\], v = u + at
Complete step-by-step solution -
As we know that the distance travelled by the body is given as 2 + 3n.
So, let the initial velocity of the body will be u metres/sec
And the acceleration of the body is a metre/\[{\sec ^2}\]
Now as we know that if u is the initial velocity and a is the acceleration of the body then distance travelled by it in n seconds is \[u + a\left( {n - \dfrac{1}{2}} \right)\]
So, \[2 + 3n = u + a\left( {n - \dfrac{1}{2}} \right)\]
\[2 + 3n = u + an - \dfrac{a}{2}\]
Now let us compare the coefficient of n and constant terms on both the sides of the above equation. We get,
3n = an (1)
And, \[2 = u - \dfrac{a}{2}\] (2)
On solving equation 1. We get,
a = 3 metre/\[{\sec ^2}\] (acceleration of the body)
So, putting the value of a in equation 2. We get,
\[2 = u - \dfrac{3}{2} = u - 1.5\]
u = 3.5 metres/sec (initial velocity of the body)
Now as we know that according to the first law of motion.
Speed at the end of t seconds is calculated as v = u + at, where u is the initial velocity and a is the acceleration.
So, velocity at the end of 2 seconds will be v = 3.5 + 3*2 = 9.5 metres/seconds
Hence, the velocity at the end of two seconds will be 9.5 m/sec.
Note:- Whenever we come up with this type of problem first, we should remember the formula that the distance travelled by a body during n seconds is \[u + a\left( {n - \dfrac{1}{2}} \right)\]. So, we compare this formula with the given distance and the coefficient of n on both the sides of the equation must be equal. So, from that we will get the value of u and a. After that we will use the first law of motion which states that the velocity at the end of t seconds is equal to v and v = u + at. This will be the easiest and efficient way to find the solution of the problem.
Formula used :- Distance travelled in n seconds = \[u + a\left( {n - \dfrac{1}{2}} \right)\], v = u + at
Complete step-by-step solution -
As we know that the distance travelled by the body is given as 2 + 3n.
So, let the initial velocity of the body will be u metres/sec
And the acceleration of the body is a metre/\[{\sec ^2}\]
Now as we know that if u is the initial velocity and a is the acceleration of the body then distance travelled by it in n seconds is \[u + a\left( {n - \dfrac{1}{2}} \right)\]
So, \[2 + 3n = u + a\left( {n - \dfrac{1}{2}} \right)\]
\[2 + 3n = u + an - \dfrac{a}{2}\]
Now let us compare the coefficient of n and constant terms on both the sides of the above equation. We get,
3n = an (1)
And, \[2 = u - \dfrac{a}{2}\] (2)
On solving equation 1. We get,
a = 3 metre/\[{\sec ^2}\] (acceleration of the body)
So, putting the value of a in equation 2. We get,
\[2 = u - \dfrac{3}{2} = u - 1.5\]
u = 3.5 metres/sec (initial velocity of the body)
Now as we know that according to the first law of motion.
Speed at the end of t seconds is calculated as v = u + at, where u is the initial velocity and a is the acceleration.
So, velocity at the end of 2 seconds will be v = 3.5 + 3*2 = 9.5 metres/seconds
Hence, the velocity at the end of two seconds will be 9.5 m/sec.
Note:- Whenever we come up with this type of problem first, we should remember the formula that the distance travelled by a body during n seconds is \[u + a\left( {n - \dfrac{1}{2}} \right)\]. So, we compare this formula with the given distance and the coefficient of n on both the sides of the equation must be equal. So, from that we will get the value of u and a. After that we will use the first law of motion which states that the velocity at the end of t seconds is equal to v and v = u + at. This will be the easiest and efficient way to find the solution of the problem.
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