JAMB 2018 Physics · Question 1

To keep a vehicle moving at a constant speed V requires power P from the engine. The force provided by the engine is

Options

  1. A.Pv\frac{P}{v}
  2. B.(12\frac{1}{2})V
  3. C.PV
  4. D.PV2\frac{P}{V^2}

Answer: A. Pv\frac{P}{v}

Explanation written with AI and checked against the answer key

1. Analyze the question

The question asks for the relationship between the power (PP) produced by a vehicle's engine, the constant speed (VV) at which the vehicle is moving, and the force (FF) exerted by the engine.

2. Explain the concept

In physics, Power is defined as the rate at which work is done or the rate at which energy is transferred. Mathematically, it is expressed as:

P=WtP = \frac{W}{t}

Since Work (WW) is equal to Force (FF) multiplied by distance (dd), we can write:

P=F×dtP = \frac{F \times d}{t}

We also know that velocity (or speed) is defined as distance divided by time (v=dtv = \frac{d}{t}). By substituting vv into the power equation, we get the fundamental relationship between Power, Force, and Velocity:

P=F×vP = F \times v
3. Show the solution

To find the force (FF), we need to rearrange the formula P=F×vP = F \times v to make FF the subject of the formula:

  1. Start with the equation: P=F×vP = F \times v
  2. Divide both sides by vv: $Pv=F×vv\frac{P}{v} = \frac{F \times v}{v}$
  3. The vv on the right side cancels out, leaving: $F=PvF = \frac{P}{v}$
4. Identify the answer

Comparing our derived formula (F=PvF = \frac{P}{v}) with the given options, it matches option A.

5. Exam Tip

A great way to remember this is the "Power-Force-Velocity" triangle. Just like the "Distance-Speed-Time" triangle, if you put PP at the top and FF and vv at the bottom, you can easily cover the variable you are looking for to see the operation needed. If you want FF, cover it, and you are left with PP over vv.

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