Questions: What is the acceleration of a car moving along a straight road that increases its speed from 0 to 100 km/h in 10 s? 1 km/h ⋅ s 100 km/h−s 10 km/h ⋅ s 10 m/s^2

Solution Steps

First, we need to convert the speed from kilometers per hour to meters per second. The conversion factor is:

\[ 1 \, \text{km/h} = \frac{1000 \, \text{m}}{3600 \, \text{s}} = \frac{5}{18} \, \text{m/s} \]

Thus, the final speed in meters per second is:

\[ 100 \, \text{km/h} \times \frac{5}{18} \, \text{m/s per km/h} = \frac{500}{18} \, \text{m/s} \approx 27.7778 \, \text{m/s} \]

Acceleration is defined as the change in velocity divided by the time taken for that change. The initial velocity \( v_i \) is \( 0 \, \text{m/s} \), the final velocity \( v_f \) is \( 27.7778 \, \text{m/s} \), and the time \( t \) is \( 10 \, \text{s} \).

The formula for acceleration \( a \) is:

\[ a = \frac{v_f - v_i}{t} = \frac{27.7778 \, \text{m/s} - 0 \, \text{m/s}}{10 \, \text{s}} = 2.7778 \, \text{m/s}^2 \]

The calculated acceleration \( 2.7778 \, \text{m/s}^2 \) does not directly match any of the given options. However, we can convert it back to the units used in the options:

\[ 2.7778 \, \text{m/s}^2 \times \frac{18}{5} \, \text{km/h per m/s} = 10 \, \text{km/h per s} \]

Final Answer