At time t = 0 a particle starts travelling from a height 7 z ^   cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3 t and 5 t 3 respectively. At t = 1   s acceleration of the particle will be

Question

At time t = 0 a particle starts travelling from a height 7 z ^ &#160; cm in a plane keeping z coordinate constant. At any instant of time, it's position along the x and y directions are defined as 3 t and 5 t 3 respectively. At t = 1 &#160; s acceleration of the particle will be

Aicharya · Accepted Answer

The position vector of the particle can be written as,
$\vec{r} = 3 t \hat{i} + 5 t^{3} \hat{j} + 7 \hat{k}$
The velocity of the particle will be,
$\vec{v} = \frac{d \vec{r}}{d t} = 3 \hat{i} + 15 t^{2} \hat{j}$
Now the acceleration of the particle will be,
$\vec{a} = \frac{d \vec{v}}{d t} = \frac{d^{2} \vec{r}}{d t^{2}} = 30 t \hat{j}$
At $t = 1 s \Rightarrow \frac{d^{2} r}{d t^{2}} = 30 \hat{j}$

Alternative Solution: 1. Key Concepts Involved: The problem involves concepts from kinematics, specifically the relations between position, velocity, and acceleration. The position of the particle is given as a function of time, from which we can compute velocity by taking derivative with respect to time, and acceleration by taking the second derivative. 2. Step-by-Step Solution: Given the position along the x and y directions as $x=3t$ and $y=5t^3$ respectively, we can find the velocity and acceleration in each direction by taking derivatives. The velocity along the x-axis (vx) is the derivative of x with respect to time: $v_x = \frac{dx}{dt} = \frac{d(3t)}{dt} = 3$. The velocity along the y-axis (vy) is the derivative of y with respect to time: $v_y = \frac{dy}{dt} = \frac{d(5t^3)}{dt} = 15t^2$. The acceleration along the x-axis (ax) is the derivative of vx with respect to time: $a_x = \frac{dv_x}{dt} = \frac{d(3)}{dt} = 0$. The acceleration along the y-axis (ay) is the derivative of vy with respect to time: $a_y = \frac{dv_y}{dt} = \frac{d(15t^2)}{dt} = 30t$. At $t=1s$, $a_x = 0$ and $a_y = 30t = 30*1 = 30$. So the acceleration of the particle is $30\hat{y}$. 3. Why the Correct Answer is Right: The correct answer is B) $30\hat{y}$ because at $t=1s$, the acceleration of the particle in the y direction is 30 and there is no acceleration in the x direction. The $\hat{y}$ notation indicates the direction of the acceleration. 4. Why the Other Options are Incorrect: Without knowing the other options, it is hard to explain why they are incorrect. However, any answer that does not give the acceleration as $30\hat{y}$ at $t=1s$ would be incorrect based on the given position functions. 5. Tips to Remember: Remember that the derivative of the position function with respect to time gives velocity

At time t = 0 a particle starts travelling from a height 7 z ^ &#160; cm in a plane keeping z coordinate…

Solution

At time t = 0 a particle starts travelling from a height 7 z ^ cm in a plane keeping z coordinate…