# Calculating Velocity and Distance with Simple Projections

• phospho
In summary, the conversation discussed the difference between velocity and position vectors, as well as the calculation of a particle's velocity vector and distance from the point of projection after 2 seconds. The velocity vector was found to be (10i - 4.6j) and the distance from the point of projection was 22.5m. The conversation also clarified that the horizontal component of the velocity vector remains the same while the vertical component decreases due to gravity.

#### phospho

A particle is projected with velocity vector (10i + 15j)ms^-1 where i and j are unit vectors in the horizontal and upward vertical directions respectively. Find its velocity vector 2 seconds later and its distance from the point of projection at this time.

Well first of all, could anyone explain the difference between the velocity vector after 2 seconds, and the position vector after 2 seconds?

working: I got 20i + 10.4j for the position vector after 2 seconds, using pythagoras I was able to find the distance from the point of projection (22.5m), for the velocity vector I didn't really know what to do, so I looked at the answer and deduced how they got it. So using the initial velocity vector (10i + 15j) as a point O, and the new position (20i + 10.4j) I manage to get the velocity vector to be (20-10)i + (15-10.4)j so 10i + 4.6j, however the answer is 10i - 4.6j

Am I wrong, or is the book? If further proof of working is needed I'll scan the paper I used.

Thanks.

You can't mix velocity and position vectors in the same equation freely. The units wouldn't make sense.

This is a particle under the influence of gravity, is it not? Every second, the particle is losing vertical velocity but not horizontal velocity. How much vertical velocity must it lose every second? How much vertical velocity does it lose after two seconds?

Sorry to say but u are wrong

ok do u know all about the projectile motion??

in projectile horizontal component(here 10i) dosent change.
So initial and final velocity of projectile along that direction will be same

now for vertical direction, there is acceleration g=9.8 m/s^2

so using Final velocity= initial velocity + Acceleration*Time we get...

v(along j axis)= 15(initial)-9.8*2 {acceleration is downward but projection is upward so a =-9.8}

Solving we get v(along j axis) = -4.6

my bad, vertically it loses v = 15 - 9.8*2 = -4.6, and horizontally it gains 10*2 = 20

thanks!

The position vector can be described as a vector from the origin of an assigned coordinate system to the point of the particle in the xy plane, defined by it's x and y coordinates (in this case). This position vector will continually change as the particle moves along some trajectory. The velocity vector is always tangential to the particle's trajectory.
So we know $v_o = 10\hat{i} + 15\hat{j}.$ I assume we are in an environment with no air resistance, so the horizontal component is fixed for the duration of the flight (as can be verified by calculation).
The vertical component of velocity at some time $t$ is given by $v_y = v_{oy} -gt.$ Use this to compute your vertical component after 2 seconds.

...horizontally it gains 10*2 = 20

No there is no acceleration along horizontal so horizontal component remains same

## 1. What is the formula for calculating velocity with simple projections?

The formula for calculating velocity with simple projections is v = d/t, where v is the velocity, d is the distance, and t is the time.

## 2. How do you calculate distance with simple projections?

To calculate distance with simple projections, you can use the formula d = vt, where d is the distance, v is the velocity, and t is the time.

## 3. What is the difference between velocity and speed?

Velocity and speed are often used interchangeably, but they have slightly different meanings. Velocity is a vector quantity that includes both speed and direction, while speed is a scalar quantity that only measures how fast an object is moving.

## 4. Can you use simple projections to calculate distance and velocity for objects with changing speeds?

Yes, simple projections can still be used to calculate distance and velocity for objects with changing speeds. However, the calculations may be more complicated if the speed is not constant and may require using calculus.

## 5. How do you incorporate acceleration into calculating velocity and distance with simple projections?

To incorporate acceleration into calculating velocity and distance with simple projections, you can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Similarly, for distance, you can use the formula d = ut + 1/2at^2.