An auto travelling along a straight road increases its speed from `30.0 m s^(-1)` to `50.0 m s^(-1)` in a distance of 180 m. If the acceleration is constant, how much time elapse while the auto moves this distance?

Options

A. 6.0 s
B. 4.5 s
C. 3.6 s
D. 7.0 s

Explanation

To solve the problem step by step, we will use the equations of motion.

### Step 1: Identify the given values
- Initial speed (u) = 30.0 m/s
- Final speed (v) = 50.0 m/s
- Distance (s) = 180 m

### Step 2: Use the equation of motion to find acceleration (a)
We can use the equation:
\[ v^2 = u^2 + 2as \]

Substituting the known values:
\[ (50.0)^2 = (30.0)^2 + 2a(180) \]

Calculating the squares:
\[ 2500 = 900 + 360a \]

### Step 3: Rearrange the equation to solve for acceleration (a)
Subtract 900 from both sides:
\[ 2500 - 900 = 360a \]
\[ 1600 = 360a \]

Now, divide both sides by 360:
\[ a = \frac{1600}{360} \]
\[ a = \frac{40}{9} \, \text{m/s}^2 \]

### Step 4: Use the acceleration to find the time (t)
Now we will use the equation:
\[ v = u + at \]

Substituting the known values:
\[ 50 = 30 + \left(\frac{40}{9}\right)t \]

### Step 5: Rearrange the equation to solve for time (t)
Subtract 30 from both sides:
\[ 50 - 30 = \left(\frac{40}{9}\right)t \]
\[ 20 = \left(\frac{40}{9}\right)t \]

Now, multiply both sides by \( \frac{9}{40} \):
\[ t = 20 \cdot \frac{9}{40} \]
\[ t = \frac{180}{40} \]
\[ t = 4.5 \, \text{seconds} \]

### Final Answer
The time elapsed while the auto moves this distance is **4.5 seconds**.

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