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A drag racer has a final velocity of 520 km/hr. If she takes 3.8 seconds to travel 400m, what is her average acceleration?

a) 137.89 m/s²
b) 145.26 m/s²
c) 129.47 m/s²
d) 132.63 m/s²

User Chronio
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1 Answer

4 votes

Final answer:

The average acceleration of the drag racer is calculated using kinematic equations. After converting the final velocity to meters per second, the calculation shows an average acceleration of approximately 38.01 m/s^2 over the time period and 52.06 m/s^2 over the distance, but these results do not match the given answer choices, suggesting an error.

Step-by-step explanation:

To find the average acceleration of the drag racer, we can use the following kinematic equation:

\(a = \frac{v_f - v_i}{t}\)

where:

  • \(a\) is the average acceleration
  • \(v_f\) is the final velocity
  • \(v_i\) is the initial velocity (which we can assume as 0 since the dragster starts from rest)
  • \(t\) is the time taken to reach the final velocity

First, we convert the final velocity from km/hr to m/s:

\(520 \frac{km}{hr} \times \frac{1000 m}{1 km} \times \frac{1 hr}{3600 s} = 144.44 \frac{m}{s}\)

Now we can calculate the average acceleration:

\(a = \frac{144.44 m/s - 0 m/s}{3.8 s} = 38.01 m/s^2\)

To find the average acceleration over the distance of 400m, we can use another kinematic equation:

\(v_f^2 = v_i^2 + 2a \times d\)

where:

  • \(v_f\) is the final velocity (144.44 m/s)
  • \(v_i\) is the initial velocity (0 m/s)
  • \(a\) is the average acceleration
  • \(d\) is the distance traveled (400m)

Solving for acceleration yields:

\(a = \frac{v_f^2 - v_i^2}{2d} = \frac{(144.44 m/s)^2}{2 \times 400 m} \approx 52.06 m/s^2\)

However, none of the provided answer choices (137.89 m/s^2, 145.26 m/s^2, 129.47 m/s^2, 132.63 m/s^2) corresponds to the calculated value. It appears there might be an error in the question or the provided answer choices.

User Marthym
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7.8k points