An object with an initial velocity of 4 m/s moves along a straight axis under constant acceleration.Three sec later its velocity is 14m/s.How far did it travel during this time?
A.27
B.17
C.67
D.57
E.77
average velocity is ... (4 + 14) / 2
d = v t
i got 3.33
To determine how far the object traveled during this time, we can use the kinematic equation:
\[v_f = v_i + at \]
Where:
- \(v_f\) is the final velocity (14 m/s)
- \(v_i\) is the initial velocity (4 m/s)
- \(a\) is the acceleration
- \(t\) is the time (3 seconds)
Rearranging the equation to solve for acceleration gives us:
\[a = \frac{{v_f - v_i}}{{t}} \]
Substituting the known values:
\[a = \frac{{14 \, \text{m/s} - 4 \, \text{m/s}}}{{3 \, \text{s}}} \]
Calculating this, we get:
\[a = \frac{{10 \, \text{m/s}}}{{3 \, \text{s}}} \]
Now that we have the acceleration, we can use another kinematic equation to determine the distance traveled:
\[d = v_i t + \frac{1}{2} a t^2 \]
Substituting the known values:
\[d = 4 \, \text{m/s} \times 3 \, \text{s} + \frac{1}{2} \left(\frac{{10 \, \text{m/s}}}{{3 \, \text{s}}}\right) \left(3 \, \text{s}\right)^2 \]
Calculating this, we get:
\[d = 12 \, \text{m} + \frac{1}{2} \times \frac{{10 \, \text{m/s}}}{{3 \, \text{s}}} \times 9 \, \text{s}^2 \]
\[d = 12 \, \text{m} + \frac{{90 \, \text{m}}}{{6 \, \text{s}}} \]
\[d = 12 \, \text{m} + 15 \, \text{m} \]
\[d = 27 \, \text{m} \]
Therefore, the object traveled 27 meters during this time. The correct answer is A. 27.