Question

A stone is thrown horizontally with the velocity 15m/s. Determine the tangential and normal accelerations of the stone in 1 second after it begins to move.

Answer 1

tangential acceleration = 5.36 m/s^2, normal acceleration = 8.20 m/s^2

Answer 2

The stone undergoes projectile motion under the influence of gravity. We need to find the tangential and normal accelerations at $t=1$ second.

Let's set up a coordinate system where the x-axis is horizontal (in the direction of initial velocity) and the y-axis is vertical upwards.

1. Velocity components at $t=1$ s: Initial horizontal velocity, $v_{x0} = 15$ m/s. Initial vertical velocity, $v_{y0} = 0$ m/s. Acceleration due to gravity, $g = 9.8$ m/s$^2$ (downwards). So, $\vec{a} = -9.8 \hat{j}$ m/s$^2$.

   At time $t=1$ s: Horizontal velocity: $v_x(t) = v_{x0} = 15$ m/s (since there is no horizontal acceleration). Vertical velocity: $v_y(t) = v_{y0} + at = 0 + (-9.8)(1) = -9.8$ m/s.

   The velocity vector at $t=1$ s is $\vec{v} = (15 \hat{i} - 9.8 \hat{j})$ m/s. The magnitude of the velocity is $v = |\vec{v}| = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + (-9.8)^2} = \sqrt{225 + 96.04} = \sqrt{321.04} \approx 17.917$ m/s.

2. Acceleration vector: The only acceleration acting on the stone is due to gravity: $\vec{a} = -9.8 \hat{j}$ m/s$^2$. The magnitude of acceleration is $a = |\vec{a}| = 9.8$ m/s$^2$.

3. Tangential Acceleration ($a_t$): Tangential acceleration is the component of the acceleration vector along the direction of the velocity vector. It determines the rate of change of speed. $a_t = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$

   First, calculate the dot product $\vec{a} \cdot \vec{v}$: $\vec{a} \cdot \vec{v} = (0 \hat{i} - 9.8 \hat{j}) \cdot (15 \hat{i} - 9.8 \hat{j})$ $\vec{a} \cdot \vec{v} = (0)(15) + (-9.8)(-9.8) = 0 + 96.04 = 96.04$

   Now, calculate $a_t$: $a_t = \frac{96.04}{17.917} \approx 5.360$ m/s$^2$.

4. Normal Acceleration ($a_n$): Normal acceleration (also known as centripetal acceleration) is the component of the acceleration vector perpendicular to the velocity vector. It causes the change in the direction of velocity. The magnitude of the total acceleration is $a = \sqrt{a_t^2 + a_n^2}$. Therefore, $a_n = \sqrt{a^2 - a_t^2}$

   $a_n = \sqrt{(9.8)^2 - (5.360)^2}$ $a_n = \sqrt{96.04 - 28.7296}$ $a_n = \sqrt{67.3104} \approx 8.204$ m/s$^2$.

The tangential acceleration is approximately $5.36$ m/s$^2$ and the normal acceleration is approximately $8.20$ m/s$^2$.

Explanation of the solution:

1. Calculate velocity components ($v_x, v_y$) at $t=1$s using kinematic equations.
2. Determine the magnitude of the resultant velocity ($v$).
3. Identify the acceleration vector ($\vec{a}$) which is due to gravity.
4. Calculate tangential acceleration ($a_t$) using the dot product of acceleration and velocity vectors: $a_t = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|}$.
5. Calculate normal acceleration ($a_n$) using the relationship $a_n = \sqrt{|\vec{a}|^2 - a_t^2}$.