a sports car accelerates from a standing start to 65 mi\h in 4.61s how far can it travel in that time
1 answer:
The car can travel 220.27 ft in 4.61 s of time.
Given that a sports car accelerates from a standing start to 65 mi/hr.
So the initial velocity = 0 ft/s
The final velocity = 65 mi/hr.
We convert the final velocity from mi/hr. to ft/s.
1 mi/hr = 5280/3600 ft/s
= 1.47 ft/s
Therefore final velocity = 65 x 1.47 = 95.55 ft/s
The time taken by the car = 4.61 s
Now the average acceleration of the car = final velocity - initial velocity / time taken = 95.55 - 0/ 4.61 s = 20.73
The displacement of the car , ,
where u is the initial velocity of the car, t, the time taken by the car and a, the average acceleration.
So, displacement = 0 x 4.61 + 1/2 x 20.73 x
= 220.27 ft
Hence the distance travelled by the sports car in 4.61s is 220.27 ft.
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Answer:
a) The distance of spectator A to the player is 79.2 m
b) The distance of spectator B to the player is 43.9 m
c) The distance between the two spectators is 90.6 m
Explanation:
a) Knowing the time it takes the sound to reach both spectators, we can calculate their position relative to the player, using this equation:
x = v * t
where:
x = position of the spectators
v = speed of sound
t = time
Then, the position for spectator A relative to the player is:
x = 343 m/s * 0.231 s = 79.2 m
b)For spectator B:
x = 343 m/s * 0.128 s
x = 43.9 m
The distance of spectator A and B to the player is 79.2 m and 43.9 m respectively.
c) To calculate the distance between the spectators, please see the attached figure. Notice that the distance between the spectators is the hypotenuse of the triangle formed by the sightline of both. We already know the longitude of the two sides. Then, using Pythagoras theorem:
(Distance AB)² = A² + B²
(Distance AB)² = (79.2 m)² + (43.9 m)²
Distance AB = 90. 6 m
