# Projectile Motion/Kinematics/Vectors

• Delta G
In summary, the robot submersible will be located 13126.8366 meters in the x-direction and 14881.2836 meters in the y-direction from the research vessel after completing the given sequence of accelerating, maintaining velocity, and coming to a full stop.
Delta G

## Homework Statement

A robot submersible is released from a research vessel. Through computer controls the craft is to execute the following sequence: a)a = 3.188i − 3.608j m/s2 for 23 s, b) maintain its velocity (no acceleration) for another 2.59 min, and c) come to a full stop. How far from the vessel will it be located?

a. 19008i - 21008j m
b. 9408i - 11008j m
c.10008i + 12008j m
d. 10008i - 12008j m

vf^2 = v0^2 +2ax

## The Attempt at a Solution

No idea where to start

Last edited:
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Hi there! I can help you with this problem. Let's break it down step by step.

First, we have the acceleration (a) and time (t) given in the problem. We can use the equation vf = v0 + at to find the final velocity (vf) after 23 seconds. The initial velocity (v0) is assumed to be 0 since the robot submersible is starting from rest. Plugging in the values, we get:

vf = (0) + (3.188i - 3.608j m/s^2)(23 s) = 73.324i - 83.064j m/s

Next, we need to find the displacement (d) of the robot submersible during the first 23 seconds. We can use the equation d = v0t + 1/2at^2, assuming that the initial displacement (d0) is also 0. Plugging in the values, we get:

d = (0)(23 s) + 1/2(3.188i - 3.608j m/s^2)(23 s)^2 = 1732.236i - 1964.652j m

Now, the robot submersible maintains its velocity for another 2.59 minutes, or 155.4 seconds. During this time, it will not experience any acceleration, so its velocity will remain constant at 73.324i - 83.064j m/s. To find the displacement during this time, we can use the equation d = vt, where v is the constant velocity and t is the time. Plugging in the values, we get:

d = (73.324i - 83.064j m/s)(155.4 s) = 11394.6096i - 12916.6416j m

Finally, the robot submersible comes to a full stop, so its final velocity (vf) is 0. We can use the equation vf^2 = v0^2 + 2ad to find the displacement during this time. Plugging in the values, we get:

0 = (73.324i - 83.064j m/s)^2 + 2a(d)
Solving for d, we get:
d = -0.009i + 0.01j m

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I would approach this problem by first identifying the relevant concepts and equations involved in projectile motion, kinematics, and vectors. Projectile motion involves the motion of an object in a curved path due to the influence of gravity. Kinematics deals with the motion of objects without considering the forces that cause the motion. Vectors are quantities that have both magnitude and direction.

The given information provides us with the acceleration (a) of the robot submersible, the time (t) for which it experiences this acceleration, and the time (t') for which it maintains its velocity. We can use the equation vf = v0 + at to find the final velocity (vf) after the acceleration phase. Since the initial velocity (v0) is not given, we can assume it to be zero.

vf = v0 + at
vf = 0 + (3.188i - 3.608j) * 23
vf = 73.324i - 83.104j m/s

Next, we can use the equation x = x0 + v0t + 1/2at^2 to find the distance (x) covered during the acceleration phase. Again, we can assume the initial position (x0) to be zero.

x = x0 + v0t + 1/2at^2
x = 0 + 0 + 1/2(3.188i - 3.608j) * (23)^2
x = 19008i - 21008j m

After the acceleration phase, the submersible maintains its velocity for another 2.59 minutes (t'). This means that it will cover a distance of v*t' = (73.324i - 83.104j) * (2.59 * 60) = 9408i - 11008j m.

Finally, to find the total distance from the vessel, we can simply add the distances covered during the acceleration and velocity-maintenance phases.

Total distance = 19008i - 21008j + 9408i - 11008j = 28416i - 32016j m

Therefore, the correct answer is option b) 9408i - 11008j m.

## 1. What is projectile motion?

Projectile motion is the motion of an object through the air due to the force of gravity. It follows a curved path known as a parabola and is influenced by the initial velocity and angle of launch.

## 2. What is the difference between kinematics and dynamics?

Kinematics is the study of the motion of objects without considering the forces that cause the motion. Dynamics, on the other hand, is the study of the forces that cause motion and their effects on objects.

## 3. How do you calculate the velocity of a projectile?

The velocity of a projectile can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time.

## 4. What is a vector quantity?

A vector quantity is a physical quantity that has both magnitude and direction. Examples include displacement, velocity, and acceleration.

## 5. How does air resistance affect projectile motion?

Air resistance, also known as drag, can affect a projectile's motion by slowing it down and changing its trajectory. This is because air resistance acts in the opposite direction of the projectile's motion, reducing its velocity and causing it to follow a slightly different path.

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