
The acceleration of a particle is $a = 3{t^2}\hat i + 5t\hat j - (8{t^3} + 400)\hat k{\text{ }}m/{s^2}$. What is the change in velocity from $t = 0$ to 10 s given as ${n_1}\hat i + {n_2}\hat j - {n_3}\hat k$?
(A) ${n_1} = 1000m/s$
(B) ${n_2} = 250m/s$
(C) ${n_3} = 24000m/s$
(D) All of these
Answer
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Hint
The acceleration of a particle is given as the rate of change of its velocity. Since velocity is a vector quantity, acceleration is also a vector, and is measurable in all the principle directions.
Complete step by step answer
In this question, we are provided with the acceleration vector of a particle. This vector tells us the acceleration of the particle in the X, Y and Z direction independently as:
$a = 3{t^2}\hat i + 5t\hat j - (8{t^3} + 400)\hat k{\text{ }}m/{s^2}$
We are required to find the velocity of the particle from
Initial time $t = 0$s
Final time $t = 10$s
Now, we know that the acceleration of a body is given as the rate of change of its velocity. Mathematically this can be written as:
$a = \dfrac{{dV}}{{dt}}$
To find the velocity, we cross-multiply and perform integration as:
$V = \int {dV} = \int {adt} $ and the limits of this integral range from 0 to 10 seconds.
Substituting the known values gives us:
$V = \int\limits_0^{10} {(3{t^2}\hat i + 5t\hat j - (8{t^3} + 400)\hat k)dt} $
Following the linearity property of the integral in the principle direction gives us:
$V = \int\limits_0^{10} {3{t^2}dt\hat i + 5tdt\hat j - (8{t^3} + 400)dt\hat k} $
Performing the integration according to the specified rules:
$V = \left| {\dfrac{{3{t^3}}}{3}\hat i + \dfrac{{5{t^2}}}{2}\hat j - (\dfrac{{8{t^4}}}{4} + 400t)\hat k} \right|_0^{10}$
Simplifying this further:
$V = \left| {{t^3}\hat i + \dfrac{{5{t^2}}}{2}\hat j - (2{t^4} + 400t)\hat k} \right|_0^{10}$
Putting the values of the initial and final time gives us:
$V = {10^3}\hat i + \dfrac{{5 \times {{10}^2}}}{2}\hat j - (2 \times {10^4} + 400 \times 10)\hat k$ [As any value multiplied by 0 will be equal to 0]
$V = 1000\hat i + 250\hat j - (20000 + 4000)\hat k$
This is equivalent to:
$V = 1000\hat i + 250\hat j - 24000\hat k$
Now comparing this vector with the general vector ${n_1}\hat i + {n_2}\hat j - {n_3}\hat k$, we get the values as:
${n_1} = 1000m/s$
${n_2} = 250m/s$
${n_3} = 24000m/s$
All of these match with the options provided, hence the correct answer is option D: All of these.
Note
The components of a vector undergo mathematical operations individually. This property is particularly helpful in determining the behaviour of a vector quantity in a certain direction. For example, if a vector is multiplied by a constant A, all of the components in all the directions will be multiplied by the same factor A.
The acceleration of a particle is given as the rate of change of its velocity. Since velocity is a vector quantity, acceleration is also a vector, and is measurable in all the principle directions.
Complete step by step answer
In this question, we are provided with the acceleration vector of a particle. This vector tells us the acceleration of the particle in the X, Y and Z direction independently as:
$a = 3{t^2}\hat i + 5t\hat j - (8{t^3} + 400)\hat k{\text{ }}m/{s^2}$
We are required to find the velocity of the particle from
Initial time $t = 0$s
Final time $t = 10$s
Now, we know that the acceleration of a body is given as the rate of change of its velocity. Mathematically this can be written as:
$a = \dfrac{{dV}}{{dt}}$
To find the velocity, we cross-multiply and perform integration as:
$V = \int {dV} = \int {adt} $ and the limits of this integral range from 0 to 10 seconds.
Substituting the known values gives us:
$V = \int\limits_0^{10} {(3{t^2}\hat i + 5t\hat j - (8{t^3} + 400)\hat k)dt} $
Following the linearity property of the integral in the principle direction gives us:
$V = \int\limits_0^{10} {3{t^2}dt\hat i + 5tdt\hat j - (8{t^3} + 400)dt\hat k} $
Performing the integration according to the specified rules:
$V = \left| {\dfrac{{3{t^3}}}{3}\hat i + \dfrac{{5{t^2}}}{2}\hat j - (\dfrac{{8{t^4}}}{4} + 400t)\hat k} \right|_0^{10}$
Simplifying this further:
$V = \left| {{t^3}\hat i + \dfrac{{5{t^2}}}{2}\hat j - (2{t^4} + 400t)\hat k} \right|_0^{10}$
Putting the values of the initial and final time gives us:
$V = {10^3}\hat i + \dfrac{{5 \times {{10}^2}}}{2}\hat j - (2 \times {10^4} + 400 \times 10)\hat k$ [As any value multiplied by 0 will be equal to 0]
$V = 1000\hat i + 250\hat j - (20000 + 4000)\hat k$
This is equivalent to:
$V = 1000\hat i + 250\hat j - 24000\hat k$
Now comparing this vector with the general vector ${n_1}\hat i + {n_2}\hat j - {n_3}\hat k$, we get the values as:
${n_1} = 1000m/s$
${n_2} = 250m/s$
${n_3} = 24000m/s$
All of these match with the options provided, hence the correct answer is option D: All of these.
Note
The components of a vector undergo mathematical operations individually. This property is particularly helpful in determining the behaviour of a vector quantity in a certain direction. For example, if a vector is multiplied by a constant A, all of the components in all the directions will be multiplied by the same factor A.
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