Answer
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Hint: To solve this question, we have to determine the linear acceleration of the ball from the given information using the second equation of motion. Then, by substituting this value of acceleration in the same equation of motion, we can find out the required displacement of the ball in the first four seconds.
Formula used: The formula used to solve this question is given by
$s = ut + \dfrac{1}{2}a{t^2}$, here $s$ is the displacement covered by a particle in the time $t$ with a constant acceleration of $a$ with an initial velocity of $u$.
Complete step by step solution:
Let the acceleration of the ball be $a$.
As the acceleration of the ball is constant, we can apply the three equations of motion. According to the question, the body covers a displacement of $1$ meter in the first $2$ seconds. According the second kinematic equation of motion, we have
$s = ut + \dfrac{1}{2}a{t^2}$
As the ball is at rest at $t = 0$ seconds, so we have $u = 0$. After the first $2$ seconds, time elapsed is $t = 2s$. Also, displacement covered by the ball during this time is $s = 1m$. Substituting these above, we get
$1 = 0\left( 2 \right) + \dfrac{1}{2}a{\left( 2 \right)^2}$
$ \Rightarrow 2a = 1$
On solving we get
$a = \dfrac{1}{2}m{s^{ - 2}}$ ………………….(1)
Now, let the displacement of the ball during the first $4$ seconds be $x$. Therefore substituting $s = x$, and $t = 4s$ in the second equation of motion, we get
$x = u\left( 4 \right) + \dfrac{1}{2}a{\left( 4 \right)^2}$
The phrase “first $4$ seconds” means that time is calculated from $t = 0$ seconds. According to the question, the ball is at rest at this instant. So we substitute $u = 0$ above to get
$x = \dfrac{1}{2}a{\left( 4 \right)^2}$
From (1)
$x = \dfrac{1}{2} \times \dfrac{1}{2} \times {\left( 4 \right)^2}$
\[ \Rightarrow x = 4m\]
Hence, the ball has rolled a displacement of $4m$ in the first four seconds.
Note: We should not worry about the rolling motion of the ball. We have been only asked to find out the linear displacement which the ball has rolled in the given amount of time. The displacement which has been found out is the displacement of the center of mass of the ball.
Formula used: The formula used to solve this question is given by
$s = ut + \dfrac{1}{2}a{t^2}$, here $s$ is the displacement covered by a particle in the time $t$ with a constant acceleration of $a$ with an initial velocity of $u$.
Complete step by step solution:
Let the acceleration of the ball be $a$.
As the acceleration of the ball is constant, we can apply the three equations of motion. According to the question, the body covers a displacement of $1$ meter in the first $2$ seconds. According the second kinematic equation of motion, we have
$s = ut + \dfrac{1}{2}a{t^2}$
As the ball is at rest at $t = 0$ seconds, so we have $u = 0$. After the first $2$ seconds, time elapsed is $t = 2s$. Also, displacement covered by the ball during this time is $s = 1m$. Substituting these above, we get
$1 = 0\left( 2 \right) + \dfrac{1}{2}a{\left( 2 \right)^2}$
$ \Rightarrow 2a = 1$
On solving we get
$a = \dfrac{1}{2}m{s^{ - 2}}$ ………………….(1)
Now, let the displacement of the ball during the first $4$ seconds be $x$. Therefore substituting $s = x$, and $t = 4s$ in the second equation of motion, we get
$x = u\left( 4 \right) + \dfrac{1}{2}a{\left( 4 \right)^2}$
The phrase “first $4$ seconds” means that time is calculated from $t = 0$ seconds. According to the question, the ball is at rest at this instant. So we substitute $u = 0$ above to get
$x = \dfrac{1}{2}a{\left( 4 \right)^2}$
From (1)
$x = \dfrac{1}{2} \times \dfrac{1}{2} \times {\left( 4 \right)^2}$
\[ \Rightarrow x = 4m\]
Hence, the ball has rolled a displacement of $4m$ in the first four seconds.
Note: We should not worry about the rolling motion of the ball. We have been only asked to find out the linear displacement which the ball has rolled in the given amount of time. The displacement which has been found out is the displacement of the center of mass of the ball.
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