Respuesta :
Its velocity when it crosses the finish line is 119.40 m/s
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = ?
Time, t = 6.70 s
Displacement, s = 1/4 mi = 0.25 mi = 400 m
Substituting
s = ut + 0.5 at²
400 = 0 x 6.70 + 0.5 x a x 6.70²
a = 17.82 m/s²
Now we have equation of motion v = u + at
Initial velocity, u = 0 m/s
Final velocity, v = ?
Time, t = 6.7 s
Acceleration, a = 17.82 m/s²
Substituting
v = u + at
v = 0 + 17.82 x 6.7
v = 119.40 m/s
Its velocity when it crosses the finish line is 119.40 m/s
The final velocity of the dragster as it crosses the finish line is 120.1m/s
Given the data in the question;
Since the dragster was initially at rest;
- Initial velocity; [tex]u = 0[/tex]
- Distance travelled; [tex]s = \frac{1}{4}miles = 0.25miles = 402.336m[/tex]
- Time taken; [tex]t = 6.70s[/tex]
Final velocity; [tex]v = \ ?[/tex]
First we determine the acceleration of the dragster, using the second equation of motion:
[tex]s = ut + \frac{1}{2}at^2[/tex]
We substitute our values into the equation
[tex]402.336m = [0 * 6.70s] + [ \frac{1}{2} * a * (6.70s)^2 ]\\\\402.336m = \frac{1}{2} * a * 44.89s^2 \\\\402.336m = 22.445s^2 * a\\\\a = \frac{402.336m}{22.445s^2}\\\\a = 17.925m/s^2[/tex]
Now, to determine the final velocity of the dragster as it crosses the finish line, we use the first equation of motion:
[tex]v = u + at[/tex]
We substitute our values into the equation
[tex]v = 0 + [ 17.925m/s^2\ *\ 6.70s ]\\\\v = 17.925m/s^2\ *\ 6.70s\\\\v = 120.1m/s[/tex]
Therefore, the final velocity of the dragster as it crosses the finish line is 120.1m/s.
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