A car starts with an initial velocity of $5 \text{ m/s}$ and accelerates at a constant rate of $3 \text{ m/s}^2$. At $t=5 \text{ s}$, a person sitting in the car throws a ball vertically upwards relative to the car with a speed of $15 \text{ m/s}$. What is the magnitude of the velocity and acceleration of the ball at $t=8 \text{ s}$? (Take $g=10 \text{ m/s}^2$)

Question

Prepp Team · Accepted Answer

Detailed Explanation: Car and Ball Motion Problem

This problem asks us to find the velocity and acceleration of a ball at a specific time ($t=8 \text{ s}$), considering it's thrown from a car that is itself moving with acceleration.

Initial Conditions Analysis

Let's break down the information given in the question:

The car starts with an initial velocity of $v_{0,\text{car}} = 5 \text{ m/s}$.
The car experiences a constant acceleration of $a_{\text{car}} = 3 \text{ m/s}^2$.
The ball is thrown at time $t_0 = 5 \text{ s}$.
The ball's initial velocity relative to the car is $15 \text{ m/s}$ directed vertically upwards.
We need to find the velocity and acceleration at $t = 8 \text{ s}$.
The acceleration due to gravity is $g = 10 \text{ m/s}^2$.

Calculating Car's Velocity at $t=5 \text{ s}$

The car is accelerating. We need to know its velocity when the ball is thrown ($t=5 \text{ s}$). This velocity determines the horizontal component of the ball's initial velocity relative to the ground.

Using the formula for velocity under constant acceleration:

$v_{\text{car}}(t) = v_{0,\text{car}} + a_{\text{car}} \cdot t$

Plugging in the values for $t=5 \text{ s}$:

$v_{\text{car}}(5 \text{ s}) = 5 \text{ m/s} + (3 \text{ m/s}^2) \cdot (5 \text{ s})$

$v_{\text{car}}(5 \text{ s}) = 5 \text{ m/s} + 15 \text{ m/s}$

$v_{\text{car}}(5 \text{ s}) = 20 \text{ m/s}$

This $20 \text{ m/s}$ is the horizontal velocity of the car at the moment the ball is thrown.

Determining Ball's Initial Velocity (Relative to Ground)

When the ball is thrown upwards relative to the car, its initial velocity relative to the ground is the vector sum of the car's velocity and the velocity relative to the car.

Horizontal Component ($u_{x,\text{ball}}$): The ball inherits the car's horizontal velocity at $t=5 \text{ s}$.
$u_{x,\text{ball}} = v_{\text{car}}(5 \text{ s}) = 20 \text{ m/s}$
Vertical Component ($u_{y,\text{ball}}$): The ball is thrown upwards relative to the car at $15 \text{ m/s}$. We'll take the upward direction as positive.
$u_{y,\text{ball}} = +15 \text{ m/s}$

So, the initial velocity vector of the ball relative to the ground at $t=5 \text{ s}$ is $\vec{u}_{\text{ball}} = (20 \hat{i} + 15 \hat{j}) \text{ m/s}$.

Motion Analysis from $t=5 \text{ s}$ to $t=8 \text{ s}$

The time elapsed since the ball was thrown is:

$\Delta t = 8 \text{ s} - 5 \text{ s} = 3 \text{ s}$

Now, let's analyze the ball's motion in the horizontal and vertical directions separately. Once thrown, the ball is a projectile.

Horizontal Motion: We neglect air resistance. The ball's horizontal velocity remains constant after it's thrown.
$v_{x,\text{ball}}(8 \text{ s}) = u_{x,\text{ball}} = 20 \text{ m/s}$
Vertical Motion: The ball is only under the influence of gravity ($a_y = -g$).
Using the kinematic equation for vertical velocity:

$v_{y,\text{ball}}(t) = u_{y,\text{ball}} + a_y \cdot t$

Substitute the values for $\Delta t = 3 \text{ s}$:

$v_{y,\text{ball}}(8 \text{ s}) = 15 \text{ m/s} + (-10 \text{ m/s}^2) \cdot (3 \text{ s})$

$v_{y,\text{ball}}(8 \text{ s}) = 15 \text{ m/s} - 30 \text{ m/s}$

$v_{y,\text{ball}}(8 \text{ s}) = -15 \text{ m/s}$

The negative sign means the ball is moving downwards at $t=8 \text{ s}$.

Calculating Velocity Magnitude at $t=8 \text{ s}$

The ball's velocity vector at $t=8 \text{ s}$ is $\vec{v}_{\text{ball}}(8 \text{ s}) = (20 \hat{i} - 15 \hat{j}) \text{ m/s}$. To find the magnitude of the velocity, we use the Pythagorean theorem:

$|\vec{v}_{\text{ball}}(8 \text{ s})| = \sqrt{(v_{x,\text{ball}})^2 + (v_{y,\text{ball}})^2}$

$|\vec{v}_{\text{ball}}(8 \text{ s})| = \sqrt{(20 \text{ m/s})^2 + (-15 \text{ m/s})^2}$

$|\vec{v}_{\text{ball}}(8 \text{ s})| = \sqrt{400 + 225} \text{ m/s}$

$|\vec{v}_{\text{ball}}(8 \text{ s})| = \sqrt{625} \text{ m/s}$

$|\vec{v}_{\text{ball}}(8 \text{ s})| = 25 \text{ m/s}$

The magnitude of the ball's velocity at $t=8 \text{ s}$ is $25 \text{ m/s}$.

Determining Acceleration Magnitude at $t=8 \text{ s}$

After the ball is thrown, it moves under the influence of gravity alone (neglecting air resistance). The acceleration due to gravity is constant.

The acceleration vector is:

$\vec{a}_{\text{ball}} = -g \hat{j} = -10 \text{ m/s}^2 \hat{j}$

The magnitude of this acceleration is:

$|\vec{a}_{\text{ball}}| = \sqrt{0^2 + (-10 \text{ m/s}^2)^2}$

$|\vec{a}_{\text{ball}}| = 10 \text{ m/s}^2$

The magnitude of the ball's acceleration at $t=8 \text{ s}$ (and indeed, at any point after being thrown) is $10 \text{ m/s}^2$.

Final Answer Summary

Based on the calculations:

The magnitude of the ball's velocity at $t=8 \text{ s}$ is $25 \text{ m/s}$.
The magnitude of the ball's acceleration at $t=8 \text{ s}$ is $10 \text{ m/s}^2$.

This corresponds to the first option provided.

Prepp Team · Answer

$25 	ext{ m/s}$, $3 	ext{ m/s}^2$

Prepp Team · Answer

$10\sqrt{13} 	ext{ m/s}$, $10 	ext{ m/s}^2$

Prepp Team · Answer

$15\sqrt{2} 	ext{ m/s}$, $10 	ext{ m/s}^2$