# Kinematics question (uniformly accelerated linear motion)

• Sean1218
In summary: So you would have d = d0 + v1(t) + 1/2a(t^2).In summary, two friends see each other in a store and are initially 50m apart. One friend walks towards the other at a constant speed of 0.50 m/s, while the second friend accelerates uniformly from rest at a rate of 1.0 m/s^2. The question asks how long it will take for the two friends to shake hands. By setting up equations for each friend's position and equating them, we can find that it will take approximately 4.5 seconds for the two friends to meet and shake hands.
Sean1218

## Homework Statement

Two friends see each other in a store and are initially 50m apart. The first friend starts walking towards the second at a constant speed of 0.50 m/s. At the same instant, the second friend accelerates uniformly from rest at a rate of 1.0 m/s^2 toward the first friend. How long before the two friends can shake hands?

v = d/t
d = vt + 1/2at^2

## The Attempt at a Solution

Distance between them = 50m
Person A (on left side):
v = 0.50 m/s (is this just v or v1 or v2? not sure if i can even use the uniform acceleration equations so it's just v?)
d = vt
d = 0.50t (1)

Person B (right side):
a = -1.0 m/s^2
v1 = 0 m/s
d = 1/2(-1)t^2 (2)

(1) = (2)
0.50t = 1/2(-1)t^2
0 = -1/2t^2 - 0.50t
0 = -1/2t(t + 1)

t = 0 or t = -1

I did something wrong since t being 0 or -1 wouldn't make sense.

Not sure what else I can do or how to deal with the distance especially since they don't necessarily meet at halfway point. Any help?

You need to take the distance between the two into account.

Yes, but I'm not sure how.

If we say that the first person is at x=0, and the distance between the two is 50 m, how can we write person 2's initial position?

at x = 50 or d2i = d1i + 50, i remember this question from last year, but can't remember where I went with this

Right, so basically for person 2, the initial position is 50 m, so you need to add that to your equation for person 2, and then equate both and solve for t.

How would I phrase the problem if I had to answer it on a test? just d = v1(t) + 1/2a(t^2) + 50 as the equation? Just asking in terms of variables and such

You could do it that way, or you could call it d0, or some other name, and just say it is the initial position at t=0.

## 1. What is kinematics in physics?

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. It involves studying concepts such as position, velocity, acceleration, and time.

## 2. What is uniformly accelerated linear motion?

Uniformly accelerated linear motion is a type of motion in which the object's velocity changes at a constant rate. This means that the object's acceleration remains constant over time. An example of this is a ball rolling down a ramp due to gravity.

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

Speed is the rate at which an object covers distance, while velocity is the rate at which an object covers distance in a specific direction. In other words, velocity includes both the speed of an object and its direction of motion.

## 4. How is acceleration calculated in uniformly accelerated linear motion?

Acceleration in uniformly accelerated linear motion can be calculated using the formula a = (vf - vi) / t, where "a" represents acceleration, "vf" represents final velocity, "vi" represents initial velocity, and "t" represents time.

## 5. What is the difference between uniform and non-uniform motion?

In uniform motion, the object's speed remains constant, while in non-uniform motion, the object's speed changes. Uniform motion can be represented by a straight line on a distance-time graph, while non-uniform motion is represented by a curved line.

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