Answer
Verified
103.5k+ views
Hint: In order to find the solution of the given question we need to know the equations of motion. After applying the equations of motion, we can finally conclude with the correct solution for the given question. First of all we need to differentiate the position with respect to time to get the velocity. After that we need to differentiate the velocity in order to get the required value of the acceleration.
Complete step by step solution:
The position of the particle in the question is given as, $x = ({t^3} - 6{t^2} - 15t + 40)m$
We know that velocity, $v = \dfrac{{dx}}{{dt}} = \dfrac{{d({t^3} - 6{t^2} - 15t + 40)}}{{dt}}$
$\therefore v = 3{t^2} - 12t - 15m{s^{ - 1}}$
Also, the acceleration of a body is given by, $a = \dfrac{{dv}}{{dt}} = \dfrac{{d(3{t^2} - 12t - 15)}}{{dt}}$
$\therefore a = 6t - 12m{s^{ - 2}}$
a) We need to find the time at which the velocity is zero.
Therefore, we can write the equation as,
$\Rightarrow v = 3{t^2} - 12t - 15 = 0$
$ \Rightarrow 3{t^2} - 15t + 3t - 15 = 0$
$ \Rightarrow 3t(t - 5) + 3(t - 5) = 0$
$ \Rightarrow (3t + 3)(t - 5) = 0$
$\therefore t = - 1$ or $t = 5$
Since, the time cannot be negative, therefore, $t = 5s$
b) Now we need to find the position at $t = 5s$
$\Rightarrow x = ({t^3} - 6{t^2} - 15t + 40)m$
$ \Rightarrow x = {5^3} - 6 \times {5^2} - 15 \times 5 + 40$
$ \Rightarrow x = 125 - 150 - 75 + 40 = - 60m$
Similarly, position at $t = 0s$
$\Rightarrow x = ({t^3} - 6{t^2} - 15t + 40)m$
$ \Rightarrow x = {0^3} - 6 \times {0^2} - 15 \times 0 + 40 = 40m$
Now, we need to find the displacement at $t = 5s$ and $t = 0s$
$\Rightarrow s = {x_5} - {x_0} = - 60 - 40 = - 100m$
c) Now, we need to find the acceleration at $t = 5s$
$\Rightarrow a = 6t - 12 = 6 \times 5 - 12$
$\therefore a = 30 - 12 = 18m{s^{ - 2}}$
Therefore, the required acceleration is $18m{s^{ - 2}}$.
Note: We define velocity of an object as the rate of change of its position with respect to reference point. It is the function of time. We define acceleration as the rate of change of velocity of the object with respect to the frame of time. It is a vector quantity which means it has magnitude and direction. When an object accelerates it means the velocity keeps changing. When velocity is zero it means that there is no acceleration.
Complete step by step solution:
The position of the particle in the question is given as, $x = ({t^3} - 6{t^2} - 15t + 40)m$
We know that velocity, $v = \dfrac{{dx}}{{dt}} = \dfrac{{d({t^3} - 6{t^2} - 15t + 40)}}{{dt}}$
$\therefore v = 3{t^2} - 12t - 15m{s^{ - 1}}$
Also, the acceleration of a body is given by, $a = \dfrac{{dv}}{{dt}} = \dfrac{{d(3{t^2} - 12t - 15)}}{{dt}}$
$\therefore a = 6t - 12m{s^{ - 2}}$
a) We need to find the time at which the velocity is zero.
Therefore, we can write the equation as,
$\Rightarrow v = 3{t^2} - 12t - 15 = 0$
$ \Rightarrow 3{t^2} - 15t + 3t - 15 = 0$
$ \Rightarrow 3t(t - 5) + 3(t - 5) = 0$
$ \Rightarrow (3t + 3)(t - 5) = 0$
$\therefore t = - 1$ or $t = 5$
Since, the time cannot be negative, therefore, $t = 5s$
b) Now we need to find the position at $t = 5s$
$\Rightarrow x = ({t^3} - 6{t^2} - 15t + 40)m$
$ \Rightarrow x = {5^3} - 6 \times {5^2} - 15 \times 5 + 40$
$ \Rightarrow x = 125 - 150 - 75 + 40 = - 60m$
Similarly, position at $t = 0s$
$\Rightarrow x = ({t^3} - 6{t^2} - 15t + 40)m$
$ \Rightarrow x = {0^3} - 6 \times {0^2} - 15 \times 0 + 40 = 40m$
Now, we need to find the displacement at $t = 5s$ and $t = 0s$
$\Rightarrow s = {x_5} - {x_0} = - 60 - 40 = - 100m$
c) Now, we need to find the acceleration at $t = 5s$
$\Rightarrow a = 6t - 12 = 6 \times 5 - 12$
$\therefore a = 30 - 12 = 18m{s^{ - 2}}$
Therefore, the required acceleration is $18m{s^{ - 2}}$.
Note: We define velocity of an object as the rate of change of its position with respect to reference point. It is the function of time. We define acceleration as the rate of change of velocity of the object with respect to the frame of time. It is a vector quantity which means it has magnitude and direction. When an object accelerates it means the velocity keeps changing. When velocity is zero it means that there is no acceleration.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main