What time and at what distance do two objects meet?

[Kuroida]

Homework Statement

I'm new at this :S

Question:
Objects A,B

A: speed = constant = 10m/s
B: starts 5.0 sec after A, starts from rest, acceleration = 1.2m/s squared

a) what time do the two objects meet? (relative to A)
b) at what distance do they meet (relative to A)

Homework Equations

The 5 kinimatic equations.
Quadratic formula

(no clue how to type them in sorry) :(

The Attempt at a Solution

Ok. I have NO clue how to start. :s

For a) you have to find delta t...right? o.0

So is it
V1=10
V2 = 10
a=0
t=?

Use 4th equation to find delta t...but that doesn't work because since a is 0 delta t gets crossed out ...o.0

What I did was for b):

a=1.2
t=5 sec
V1 = 0
d = ?

Use equation 1 (delta d = V1 x delta t +1/2a(delta t) squared)
Sub the numbers in: d= 0x5 + 1/2(1.2)(5) squared
Answer = 15m (?)

My teacher said you needed the quadradic formula though...and really I have no clue what I'm doing -_-

 

[Runaway]

Equations of motion for objects:
(object a)Uniform motion: X=vt, (Object B)-Uniform Acceleration: X= 1/2at^2
Since object A start's 5 seconds before object B, object a's time should be t+5. Using this you should be able to set the two equations equal to one another using x, and then solve for t. Then you can take that time and substitute it into either equation to find the distance at which they meet.

 

[Kuroida]

thanks. I think I got it by now

Just another question.

When I use the quadratic formula to solve for t ...

My equation is -0.6t^2 -10t -50.

t= 20.69

Do I add 5 to this number to make up for the time?

 

[Runaway]

Yes, but when you do part b, use 20.69 for t and use one of the equations you set equal to one another. you can use 25.69, but you would have to modify the equations to x=vt and x=1/2a(t-5)^2.

 

[rwisz]

Thank you for your question. I understand your confusion and am happy to help you find the answers you are looking for. First, let's break down the problem and review the relevant equations.

We have two objects, A and B. Object A has a constant speed of 10m/s and object B starts 5 seconds after object A with an acceleration of 1.2m/s squared. We are asked to find the time and distance at which these two objects meet, relative to object A.

To solve this problem, we can use the kinematic equations, specifically the equations for constant acceleration. These equations relate the variables of displacement (d), initial velocity (v0), final velocity (v), acceleration (a), and time (t). For this problem, we can use the following equations:

1. v = v0 + at
2. d = v0t + 1/2at^2
3. v^2 = v0^2 + 2ad
4. d = (v + v0)/2 * t

We also have the quadratic formula, which can be used to solve for t when given an equation in the form of ax^2 + bx + c = 0. In this problem, we will not need to use the quadratic formula.

To answer part a) of the question, we need to find the time at which the two objects meet, relative to object A. This means we are looking for the time it takes for object B to catch up to object A. We can use equation 1 to solve for t:

v = v0 + at
10 = 0 + 1.2t
t = 10/1.2 = 8.33 seconds

Therefore, object B will meet object A after 8.33 seconds.

For part b), we are asked to find the distance at which the two objects meet, relative to object A. This means we are looking for the displacement of object B from its starting point. We can use equation 2 to solve for d:

d = v0t + 1/2at^2
d = 0(8.33) + 1/2(1.2)(8.33)^2
d = 35.2 meters

Therefore, object B will meet object A at a distance of 35.2 meters, relative to object A