Respuesta :
Answer:
Its speed would be 374.35 miles per hour
Explanation:
For the average acceleration we use 2nd eqn of motion:
s = (Vo)t + at²
Vo = 0 m/s (since, car is initially at rest)
s = 400 m
t = 6.8 s
Therefore, eqn. becomes:
400 m = (0 m/s)(6.8 s) + a (6.8 s)²
a = 400 m/(6.8 s)²
a = 8.65 m/s²
Now, for a distance of 1.6 km = 1600 m, with same acceleration, its final speed can be found out by 3rd eqn. of motion:
2as = Vf² - Vo²
2(8.65 m/s²)(1600 m) = Vf² - (0 m/s)²
Vf = √27681.66
Vf = 166.38 m/s
Vf = (166.38 m/s)(1 mile/1600 m)(3600 s/ 1 hr)
Vf = 374.35 miles per hour
The distance traveled by the dragster if the race continues till 1.6 km is 526.3234 miles/h.
Given to us
The distance traveled by the dragster, s = 400 m
Time taken by the dragster to complete the race, t = 6.8 s
What is the acceleration of the dragster?
The acceleration of the dragster can be found using the second equation of motion,
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
When the race starts the initial velocity of the dragster was 0, as the dragster was at rest, therefore, u =0
substitute the value,
[tex]400=(0\times 6.8)+\dfrac{1}{2}\times a\times (6.8)^2[/tex]
[tex]a= \dfrac{400\times 2}{6.8^2}[/tex]
a = 17.3 m/s²
Thus, the acceleration of the dragster is 17.3 m/s².
What would be the velocity of the dragster if he traveled 1.6 km?
We can find out the final velocity of the dragster using the third equation of motion,
[tex]v^2-u^2 = 2as[/tex]
Substitute the values,
[tex]v^2-(0)^2 = 2(17.3)(1600)[/tex]
v = 235.287 m/s
Converting it to miles per hour
[tex]v = 235.287 \times \dfrac{\dfrac{1}{1609.34}}{\dfrac{1}{3600}} = 526.3234\rm\ miles/h[/tex]
Hence, the distance traveled by the dragster if the race continues till 1.6 km is 526.3234 miles/h.
Learn more about the Equation of motion:
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