A truck of mass 800 kg generates a power of 20000 W. How much time does the truck need to accelerate from a speed of 20 m `s^(-1)` to 30 m `s^(-1)`?

Options

A. 5 s
B. 7.5 s
C. 10 s
D. 6.3 s

Explanation

To solve the problem step by step, we will use the relationship between power, work, and kinetic energy.

### Step 1: Understand the relationship between power, work, and time
Power (P) is defined as the rate at which work (W) is done or energy is transferred. The formula is:
\[ P = \frac{W}{t} \]
Where:
- \( P \) = Power in watts (W)
- \( W \) = Work done in joules (J)
- \( t \) = Time in seconds (s)

### Step 2: Express work in terms of kinetic energy
The work done on the truck will result in a change in its kinetic energy (KE). The change in kinetic energy can be expressed as:
\[ \Delta KE = KE_{final} - KE_{initial} \]
Where:
- \( KE = \frac{1}{2} mv^2 \)
- \( m \) = mass of the truck (800 kg)
- \( v_{final} = 30 \, m/s \)
- \( v_{initial} = 20 \, m/s \)

### Step 3: Calculate the change in kinetic energy
Using the formula for kinetic energy:
\[ KE_{final} = \frac{1}{2} m v_{final}^2 = \frac{1}{2} \times 800 \times (30)^2 \]
\[ KE_{initial} = \frac{1}{2} m v_{initial}^2 = \frac{1}{2} \times 800 \times (20)^2 \]

Now, calculate each:
\[ KE_{final} = \frac{1}{2} \times 800 \times 900 = 360000 \, J \]
\[ KE_{initial} = \frac{1}{2} \times 800 \times 400 = 160000 \, J \]

Now find the change in kinetic energy:
\[ \Delta KE = KE_{final} - KE_{initial} = 360000 - 160000 = 200000 \, J \]

### Step 4: Relate power and time to the change in kinetic energy
From the power formula, we can express work done (or change in kinetic energy) in terms of power and time:
\[ W = P \times t \]
Thus,
\[ \Delta KE = P \times t \]
Substituting the values we have:
\[ 200000 = 20000 \times t \]

### Step 5: Solve for time (t)
Rearranging the equation to solve for \( t \):
\[ t = \frac{200000}{20000} = 10 \, s \]

### Final Answer
The time required for the truck to accelerate from 20 m/s to 30 m/s is **10 seconds**.

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