# 3.4 Motion with constant acceleration  (Page 7/10)

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## Strategy

We use the set of equations for constant acceleration to solve this problem. Since there are two objects in motion, we have separate equations of motion describing each animal. But what links the equations is a common parameter that has the same value for each animal. If we look at the problem closely, it is clear the common parameter to each animal is their position x at a later time t . Since they both start at ${x}_{0}=0$ , their displacements are the same at a later time t , when the cheetah catches up with the gazelle. If we pick the equation of motion that solves for the displacement for each animal, we can then set the equations equal to each other and solve for the unknown, which is time.

## Solution

1. Equation for the gazelle: The gazelle has a constant velocity, which is its average velocity, since it is not accelerating. Therefore, we use [link] with ${x}_{0}=0$ :
$x={x}_{0}+\stackrel{\text{–}}{v}t=\stackrel{\text{–}}{v}t.$

Equation for the cheetah: The cheetah is accelerating from rest, so we use [link] with ${x}_{0}=0$ and ${v}_{0}=0$ :
$x={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2}=\frac{1}{2}a{t}^{2}.$

Now we have an equation of motion for each animal with a common parameter, which can be eliminated to find the solution. In this case, we solve for t :
$\begin{array}{}\\ \\ x=\stackrel{\text{–}}{v}t=\frac{1}{2}a{t}^{2}\hfill \\ t=\frac{2\stackrel{\text{–}}{v}}{a}.\hfill \end{array}$

The gazelle has a constant velocity of 10 m/s, which is its average velocity. The acceleration of the cheetah is 4 m/s 2 . Evaluating t , the time for the cheetah to reach the gazelle, we have
$t=\frac{2\stackrel{\text{–}}{v}}{a}=\frac{2\left(10\right)}{4}=5\phantom{\rule{0.2em}{0ex}}\text{s}\text{.}$
2. To get the displacement, we use either the equation of motion for the cheetah or the gazelle, since they should both give the same answer.
Displacement of the cheetah:
$x=\frac{1}{2}a{t}^{2}=\frac{1}{2}\left(4\right){\left(5\right)}^{2}=50\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$

Displacement of the gazelle:
$x=\stackrel{\text{–}}{v}t=10\left(5\right)=50\phantom{\rule{0.2em}{0ex}}\text{m}\text{.}$

We see that both displacements are equal, as expected.

## Significance

It is important to analyze the motion of each object and to use the appropriate kinematic equations to describe the individual motion. It is also important to have a good visual perspective of the two-body pursuit problem to see the common parameter that links the motion of both objects.

Check Your Understanding A bicycle has a constant velocity of 10 m/s. A person starts from rest and runs to catch up to the bicycle in 30 s. What is the acceleration of the person?

$a=\frac{2}{3}{\phantom{\rule{0.2em}{0ex}}\text{m/s}}^{2}$ .

## Summary

• When analyzing one-dimensional motion with constant acceleration, identify the known quantities and choose the appropriate equations to solve for the unknowns. Either one or two of the kinematic equations are needed to solve for the unknowns, depending on the known and unknown quantities.
• Two-body pursuit problems always require two equations to be solved simultaneously for the unknowns.

## Conceptual questions

When analyzing the motion of a single object, what is the required number of known physical variables that are needed to solve for the unknown quantities using the kinematic equations?

State two scenarios of the kinematics of single object where three known quantities require two kinematic equations to solve for the unknowns.

If the acceleration, time, and displacement are the knowns, and the initial and final velocities are the unknowns, then two kinematic equations must be solved simultaneously. Also if the final velocity, time, and displacement are the knowns then two kinematic equations must be solved for the initial velocity and acceleration.

A spring with 50g mass suspended from it,has its length extended by 7.8cm 1.1 determine the spring constant? 1.2 it is observed that the length of the spring decreases by 4.7cm,from its original length, when a toy is place on top of it. what is the mass of the toy?
solution mass = 50g= 0.05kg force= 50 x 10= 500N extension= 7.8cm = 0.078m using the formula Force= Ke K = force/extension 500/.078 = 6410.25N/m
Sampson
1.2 Decrease in length= -4.7cm =-0.047m mass=? acceleration due to gravity= 10 force = K x e force= mass x acceleration m x a = K x e mass = K x e/acceleration = 6410.25 x 0.047/10 = 30.13kg
Sampson
1.1 6.28Nm-¹
Anita
1.2 0.03kg or 30g
Anita
I used g=9.8ms-²
Anita
you should explain how yoy got the answer Anita
Grant
ok
Anita
with the fomular F=mg I got the value for force because now the force acting on the spring is the weight of the object and also you have to convert from grams to kilograms and cm to meter
Anita
so the spring constant K=F/e where F is force and e is extension
Anita
In this first example why didn't we use P=P° + ¶hg where ¶ is density
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Hlehle
Hlehle
sorry I had a little typo in my question
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Density = m/v (mass/volume) simple as that
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