When working at a constant power of 2.5 kW against a resistance proportional to the square of its speed, the maximum speed a vehicle can attain on a level road is 50 $ms^{-1}$ .
If the vehicle accelerates from rest under the same conditions, how far does it travel before it attains half the maximum speed? The mass of the vehicle is 1500 kg.

Question

When working at a constant power of 2.5 kW against a resistance proportional to the square of its speed, the maximum speed a vehicle can attain on a level road is 50 [latex]ms^{-1}[/latex].
If the vehicle accelerates from rest under the same conditions, how far does it travel before it attains half the maximum speed? The mass of the vehicle is 1500 kg.

Accepted Answer

The power is 2500 W = force × velocity, so the engine’s driving force has a magnitude of [latex]\frac{2500}{velocity}.[/latex]

When the speed is v [latex]ms^{-1}[/latex] and the resistance is kv² N, Newton’s second law gives

[latex]\frac{2500}{v} - kv^{2} = 1500a.[/latex] ①

The maximum speed of 50 [latex]ms^{-1}[/latex] occurs when the acceleration is zero. So

[latex]\frac{2500}{50} = k(50)^{2}[/latex]

⇒ [latex] k = \frac{1}{50}.[/latex]

Substituting this value of k in ① and using [latex]a = v\frac{dv}{dx}[/latex] since x is required gives

[latex]\frac{2500}{v} - \frac{v^{2}}{50} = 1500v\frac{dv}{dx}[/latex]

⇒ 125 000 - v³ = 75 000v²[latex]\frac{dv}{dx}[/latex].

This can be solved by separating the variables and integrating.

[latex]\begin{matrix} \int{dx = \frac{75 000v^{2}}{125 000 - v^{3}}}dv & \boxed{ ext{This integral has been evaluated by inspection but could be found by substituting } z = 125000 - v^{3}.} \end{matrix} [/latex]

⇒ [latex]x = \frac{75 000}{3}ln(125 000 - v^{3}) + c[/latex]

x = 0 when v = 0 ⇒ c = 25 000 ln 125 000

so x = 25 000 ln [latex]\left(\frac{125 000}{125 000 - v^{3}} ight)[/latex]

when v = 25, x = 3338.

The distance is therefore approximately 3.34 km.

Question 18.7: When working at a constant power of 2.5 kW against a resista...

Related Answered Questions

A heavy ball is retarded by air resistance which is modelled by kv², where v ms^-1 is the speed of the ball and k is a positive constant. When falling under gravity in air, its terminal velocity is V ms^-1. i) Write an expression for the total force on the ball when it is ...

Verified Answer:

A raindrop of mass m starts from rest and falls vertically. When falling with velocity v, it experiences a resistive force of magnitude kv. i) Express the terminal velocity VT of the raindrop in terms of m, k and g. ii) Show that, if the raindrop starts from rest, its velocity v after time t is ...

Verified Answer:

Verified Answer:

Verified Answer:

A particle of mass 1 kg moves along a line with a velocity v ms^-1 under the influence of a resistive force of magnitude kv² N, where k is a constant. Initially, the velocity of the particle is 10 ms^-1 and the force continues to act until the particle has slowed down to 5 ms^-1. i) Use kinetic ...

Verified Answer:

A body of mass 2 kg, initially at rest on a smooth horizontal plane, is subjected to a horizontal force of magnitude 1/2v + 1 N, where v is the velocity of the body (v > 0). i) Find the time when the velocity is 1 ms^-1. ii) Find the displacement when the velocity is 1 ms^-1. ...

Verified Answer:

Verified Answer:

A crate of mass m is freely suspended at rest from a crane. When the operator begins to lift the crate further, the tension in the suspending cable increases uniformly from mg newtons to 1.2mg newtons over a period of 2 seconds. i) What is the tension in the cable t seconds after the lifting has ...

Verified Answer: