
The acceleration \[a\] in \[m/{s^2}\] of a particle is given by \[a = 3{t^2} + 2t + 2\] where t is the time. If the particle starts out with a velocity \[u = 2{\text{ }}m/s\] at t=0, then the velocity at the end of \[2\] second is -
\[(A)12m/s\]
\[(B)18m/s\]
\[(C)27m/s\]
\[(D)36m/s\]
Answer
516.6k+ views
Hint: The rate at which an object's velocity changes with respect to time is called acceleration. Accelerations are quantities that are measured in the vector which have both magnitude and direction. The orientation of the net force acting on an object determines the orientation of its acceleration.
The first derivative of a function that gives the position of something as a function of time is the velocity, and the second derivative is the acceleration. As a result, you differentiate position to obtain velocity, and velocity to obtain acceleration.
Complete step-by-step solution:
Since, differentiation of acceleration gives velocity. So we need to do integration for vice versa.
Now we integrate acceleration to obtain velocity,
\[v = \int {a.dt} \]
\[v(t) = {t^3} + {t^2} + 2t + c\]
Given \[v\left( 0 \right) = 2m/s\]
\[v(0) = {0^3} + {0^2} + 2 \times 0 + c\]
\[C = 2{\text{ }}m/s\]
Substituting the value of $c$ ,
\[v(t) = {t^3} + {t^2} + 2t + 2\]
\[ \Rightarrow v(2) = {2^3} + {2^2} + 2 \times 2 + 2\]
\[ \Rightarrow v(2)\; = 18m/s\]
Therefore, the velocity at the end of \[2s\] is \[18m/s.\]
Note: Acceleration is the rate of change of the velocity of an object with respect to time. Acceleration is a vector quantity that has both magnitude and direction. The orientation of an object's acceleration is represented by the orientation of the net force acting on that object. The average velocity of an object is its total displacement divided by the entire time taken. In other words, it's the speed at which an object changes its position from one place to a different place. Average velocity may be a vector quantity. The SI unit is meters per second. However, any distance unit per any unit of time can often be used when necessary, like miles per hour \[\left( {mph} \right)\] or kilometer per hour\[\left( {kmph} \right)\] .
The first derivative of a function that gives the position of something as a function of time is the velocity, and the second derivative is the acceleration. As a result, you differentiate position to obtain velocity, and velocity to obtain acceleration.
Complete step-by-step solution:
Since, differentiation of acceleration gives velocity. So we need to do integration for vice versa.
Now we integrate acceleration to obtain velocity,
\[v = \int {a.dt} \]
\[v(t) = {t^3} + {t^2} + 2t + c\]
Given \[v\left( 0 \right) = 2m/s\]
\[v(0) = {0^3} + {0^2} + 2 \times 0 + c\]
\[C = 2{\text{ }}m/s\]
Substituting the value of $c$ ,
\[v(t) = {t^3} + {t^2} + 2t + 2\]
\[ \Rightarrow v(2) = {2^3} + {2^2} + 2 \times 2 + 2\]
\[ \Rightarrow v(2)\; = 18m/s\]
Therefore, the velocity at the end of \[2s\] is \[18m/s.\]
Note: Acceleration is the rate of change of the velocity of an object with respect to time. Acceleration is a vector quantity that has both magnitude and direction. The orientation of an object's acceleration is represented by the orientation of the net force acting on that object. The average velocity of an object is its total displacement divided by the entire time taken. In other words, it's the speed at which an object changes its position from one place to a different place. Average velocity may be a vector quantity. The SI unit is meters per second. However, any distance unit per any unit of time can often be used when necessary, like miles per hour \[\left( {mph} \right)\] or kilometer per hour\[\left( {kmph} \right)\] .
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