The average speed of an object over a given time duration is 6m/s. If it travels first half of the duration with a constant speed of 4m/s, its speed during the next half duration is:
$\eqalign{
& A.{\text{ }}5m/s \cr
& B.{\text{ }}4m/s \cr
& C.{\text{ }}6m/s \cr
& D.{\text{ }}8m/s \cr} $
Answer
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Hint: The average velocity can be calculated by the sum of all the given velocities to the total number of the velocities. Average speed is given in question, so we have to calculate the velocity during the second half duration.
Complete answer:
We know that the average speed of an object is the change in the position of that object with respect to the time. According to the question the average speed of an object is 6m/s. in the first half duration, the speed of the object is 4m/s. We have to calculate the speed of the object during the next half duration. Consider using the formula of average speed.
${V_{av}} = \dfrac{{{V_1} + {V_2}}}{2}$
Where, ${V_{av}}$ is representing the average speed, ${V_1}$ is representing the speed of the object during the first half, ${V_2}$ is representing the speed of the object during the next half.
Now putting the vales ${V_1} = 4m/s,$ and ${V_{av}} = 6m/s$in above formula we get,
$\eqalign{
& \Rightarrow 6 = \dfrac{{4 + {V_2}}}{2} \cr
& \Rightarrow 12 = 4 + {V_2} \cr
& \Rightarrow {V_2} = 12 - 4 \cr
& \therefore {V_2} = 8m/s \cr} $
The speed of the object during the next half duration is $8m/s$, so the correct option among the given options is D.
Additional Information:
The average velocity of a particle in a time interval ${t_1}$ to ${t_2}$ is defined as its displacement divided by the time interval. If the particle is at a point A and time $t = {t_1}$ and at B time $t = {t_2}$, the displacement in this time interval is the vector $\overrightarrow {AB} $. Then the average velocity in this time interval is then,
$\overrightarrow {{v_{av}}} = \dfrac{{\overrightarrow {AB} }}{{{t_2} - {t_1}}}$
Note:
The average speed is the rate at which an object changes its position from one place to another place. The S.I unit of velocity is meter per second$\left( {m{s^{ - 1}}} \right)$. It can also be defined as the total distance travelled by the body in total time. It is a scalar quantity.
Complete answer:
We know that the average speed of an object is the change in the position of that object with respect to the time. According to the question the average speed of an object is 6m/s. in the first half duration, the speed of the object is 4m/s. We have to calculate the speed of the object during the next half duration. Consider using the formula of average speed.
${V_{av}} = \dfrac{{{V_1} + {V_2}}}{2}$
Where, ${V_{av}}$ is representing the average speed, ${V_1}$ is representing the speed of the object during the first half, ${V_2}$ is representing the speed of the object during the next half.
Now putting the vales ${V_1} = 4m/s,$ and ${V_{av}} = 6m/s$in above formula we get,
$\eqalign{
& \Rightarrow 6 = \dfrac{{4 + {V_2}}}{2} \cr
& \Rightarrow 12 = 4 + {V_2} \cr
& \Rightarrow {V_2} = 12 - 4 \cr
& \therefore {V_2} = 8m/s \cr} $
The speed of the object during the next half duration is $8m/s$, so the correct option among the given options is D.
Additional Information:
The average velocity of a particle in a time interval ${t_1}$ to ${t_2}$ is defined as its displacement divided by the time interval. If the particle is at a point A and time $t = {t_1}$ and at B time $t = {t_2}$, the displacement in this time interval is the vector $\overrightarrow {AB} $. Then the average velocity in this time interval is then,
$\overrightarrow {{v_{av}}} = \dfrac{{\overrightarrow {AB} }}{{{t_2} - {t_1}}}$
Note:
The average speed is the rate at which an object changes its position from one place to another place. The S.I unit of velocity is meter per second$\left( {m{s^{ - 1}}} \right)$. It can also be defined as the total distance travelled by the body in total time. It is a scalar quantity.
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