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Starting from rest, a car accelerates uniformly at 2.5 m/s². How much distance does it travel to attain a speed of 180 km/h in the direction of acceleration?

A. 5 km
B. 100 m
C. 10 km
D. 500 m


Sagot :

Let's solve this problem step by step to determine the distance a car travels to attain a speed of 180 km/h when it starts from rest and accelerates uniformly at a rate of 2.5 m/s².

### Step-by-Step Solution

Step 1: Convert the final speed from km/h to m/s
- The final speed given is 180 km/h.
- To convert this to meters per second (m/s), we use the conversion factor where 1 km/h = 5/18 m/s.

[tex]\( 180 \, \text{km/h} \times \frac{5}{18} \, \text{m/s per km/h} = 50 \, \text{m/s} \)[/tex]

Step 2: Identify the known values
- Initial speed (u) = 0 m/s (since the car is starting from rest)
- Final speed (v) = 50 m/s (as calculated in step 1)
- Acceleration (a) = 2.5 m/s²

Step 3: Use the kinematic equation to find the distance (s)
The kinematic equation that relates initial velocity (u), final velocity (v), acceleration (a), and distance (s) is:

[tex]\( v^2 = u^2 + 2as \)[/tex]

Since we need to find the distance (s), we rearrange the formula to solve for s:

[tex]\( s = \frac{v^2 - u^2}{2a} \)[/tex]

Step 4: Substitute the known values into the equation
- [tex]\( u = 0 \, \text{m/s} \)[/tex]
- [tex]\( v = 50 \, \text{m/s} \)[/tex]
- [tex]\( a = 2.5 \, \text{m/s}^2 \)[/tex]

[tex]\( s = \frac{(50 \, \text{m/s})^2 - (0 \, \text{m/s})^2}{2 \times 2.5 \, \text{m/s}^2} \)[/tex]

[tex]\( s = \frac{2500 \, \text{m}^2/\text{s}^2}{5 \, \text{m/s}^2} \)[/tex]

[tex]\( s = 500 \, \text{m} \)[/tex]

So, the distance the car should travel to attain a speed of 180 km/h is 500 meters.

Final Answer: D. 500 m