## A car and the stopping force directly proportional to its speed

### Task number: 277

A car of the mass *m* = 1000 kg is going in a straight line at a uniform velocity *v*_{0} = 30 m·s^{-1} on a horizontal road.
At a certain time *t*_{0} = 0 s a stopping force *F*_{b} starts to act against it. The stopping force is directly proportional to the speed of the car. The proportionality constant *k* = 231 kg·s^{-1}.

1) Find how the speed of the car *v*(*t*) and the position of the car *x*(*t*) vary with time.

2) Find the time at which the speed is half its initial value, i.e. 15 m·s^{-1}.

**Note:** Instead of the magnitude of the velocity we further use only the speed.

#### Hint 1 – The equation of motion

Think about the forces acting on the car and write down the equation of motion for the car.

#### Hint 2 – the stopping force and the speed

*v*(*t*)Express the stopping force

*F*_{b}from (2) using the speed of the car and solve the resulting differential equation.#### Hint 3 - the position of the car

*x(t)*The dependence of the speed on time is already known. The dependence of the position on time can be obtained by integration.

#### Hint 4 - the time at which the speed of the car is half its initial value

You are looking for the time when the speed is equal to \(\frac{v_0}{2}\). You know how the speed depends on time.

#### Complete solution

**We draw the acting forces into a figure and write the equation of motion for the car.****The following forces act on the car:**\(\vec{F}_{b}\)...the stopping force

\(\vec{F}_{G}\)...the weight

\(\vec{N}\)...the normal force exerted on the car by the surface

**Note:**The surface acts on the wheels of the car, we draw its resultant force into the figure.**The equation of motion for the car is:**\[\vec{F}_{b}+\vec{F}_{G}+\vec{N}= m\vec{a}\tag{1}\]

We choose the coordinates following the figure and write the equation (1) in scalar quantities:

\[x:\hspace{20px}-F_b = ma\tag{2}\]

\[y:\hspace{20px}N - F_G = 0\tag{3}\]

**Note:**The figure explains why there is a minus sign on the left hand side of the equation (2). The stopping force acts against the direction of the car’s motion. The forces \(\vec{F}_{G}\) and \(\vec{N}\) are perpendicular to the stopping force \(\vec{F}_{b}\) They have the same magnitude and opposite directions thus their net force is zero and the acceleration of the car in the direction of the axis*y*is zero, too.**The expression for the stopping force:**The stopping force

\[F_{b}=kv\tag{4}\]*F*_{b}is directly proportional to the speed of the car and it can be expressed by the relation:Inserting it into the equation (2) we obtain:

\[ma=-kv\tag{5}\]We express the acceleration as a change of the speed in time:

\[m\frac{dv}{dt}=-kv\tag{6}\]**The speed***v*(t)**:**We obtained the first-order differential equation with constant coefficients. We can solve it by a separation of variables.

We rewrite the equation (6) as:

\[\frac{dv}{v}=-\frac{k}{m}dt\]

and integrate both sides of it:

\[ln|v|=-\frac{k}{m}t+C\]

C is a constant of integration.

We take the inverse logarithm of both sides:

\[|v|=e^{-\frac{k}{m}t+C}=e^{-\frac{k}{m}t}\cdot e^{C}\]

The speed is positive thus we do not need to write the symbol of the absolute value. We will denote the constant

*e*^{C}by*K*.\[v=K e^{-\frac{k}{m}t }\ \]

We find the value of the constant

*K*from the initial conditions. At the time*t*= 0 s the speed was equal to*v*_{0}. Thus:\[v_0=K e^{-\frac{k}{m}\cdot 0}\ \]

and hence:

\[v_0=K\]

Finally, we get the speed as a function of time:

\[v(t)=v_0\cdot e^{-\frac{k}{m}t}\ \tag{7}\]

We know the dependence of the speed on time.

**We obtain the dependence of the position on time by integration:**\[x(t)=\int{v(t)}\,dt\]

\[x(t)=\int{v_0e^{-\frac{k}{m} t}}\,dt\]

\[x(t)=-\frac{mv_0}{k}e^{-\frac{k}{m} t}+C\]

The value of the constant

*C*can be found from the initial conditions. At the time*t*= 0 s the position of the car was equal to*x*(0) = 0 m.Thus:

\[0=-\frac{mv_0}{k}e^{-\frac{k}{m}\cdot 0}+C\]

\[C=\frac{mv_0}{k}\]

The dependence of the car’s position on time is:

\[x(t)=-\frac{mv_0}{k}e^{-\frac{k}{m}\cdot t}+\frac{mv_0}{k}\]

\[x(t)=\frac{mv_0}{k}(1-e^{-\frac{k}{m}t})\tag{8}\]

**The time at which the speed of the car is half its initial value:**We insert \(v(t)=\frac{v_0}{2}\) to the expression (7):

\[\frac{v_0}{2}=v_0e^{-\frac{k}{m}t}\ \] \[\frac{1}{2}=e^{-\frac{k}{m}t}\ \]We take the inverse logarithm of both sides:

\[ln\frac{1}{2}=-\frac{k}{m}t\]

\[-\frac{m}{k}ln\frac{1}{2}=t\]

\[\frac{m}{k}ln2=t\tag{9}\]

We insert the numerical values into (9):

\[t=\frac{1000}{231}\cdot 0.693=\frac{693}{231}=3\,\textrm{s}\]

#### Complete answer

For the dependence of the speed on time it holds \[v(t)=v_0 e^{-\frac{k}{m}t}\] .

For the dependence of the position on time it holds \[x(t)=\frac{mv_0}{k}(1-e^{-\frac{k}{m}t})\] .

The speed of the car is half its original value in 3 s.