Johnston Fuentes
09/15/2023 · Primary School
Q2. A car accelerates at a constant rate from 0 to \( 25 \mathrm{~m} / \mathrm{s} \) over a distance of 25 m . Approximately how long does it take the car to reach the velocity of \( 25 \mathrm{~m} / \mathrm{s} \) ? \( \begin{array}{llll}\text { A. } 1 \mathrm{~s} & \text { B. } 2 \mathrm{~s} & \text { C. } 4 \mathrm{~s} & \text { D. } 8 \mathrm{~s}\end{array} \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
To find the time it takes for the car to reach a velocity of \( 25 \, \text{m/s} \) while accelerating over a distance of \( 25 \, \text{m} \), we can use the kinematic equation:
\[
v^2 = u^2 + 2a s
\]
where:
- \( v \) is the final velocity (\( 25 \, \text{m/s} \)),
- \( u \) is the initial velocity (\( 0 \, \text{m/s} \)),
- \( a \) is the acceleration,
- \( s \) is the distance (\( 25 \, \text{m} \)).
Substituting the known values into the equation:
\[
(25)^2 = (0)^2 + 2a(25)
\]
This simplifies to:
\[
625 = 50a
\]
Now, solving for \( a \):
\[
a = \frac{625}{50} = 12.5 \, \text{m/s}^2
\]
Next, we can use another kinematic equation to find the time \( t \):
\[
v = u + at
\]
Substituting the known values:
\[
25 = 0 + (12.5)t
\]
Solving for \( t \):
\[
t = \frac{25}{12.5} = 2 \, \text{s}
\]
Thus, the time it takes for the car to reach a velocity of \( 25 \, \text{m/s} \) is approximately \( 2 \, \text{s} \).
The answer is **B. 2 s**.
Quick Answer
The time it takes for the car to reach \( 25 \, \text{m/s} \) is approximately \( 2 \, \text{s} \).
Therefore, the correct answer is B. 2 s.
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