The car travels a distance of 70.2 meters while accelerating from 16 m/s to 45m/s in 2.3 seconds.
Given in the question
Initial velocity of the car = 16 m/s
Final velocity of the car = 45 m/s
Initial time reading = 0 seconds
Final time reading = 2.3 seconds
Time taken = Final time reading - Initial time reading
= 2.3 - 0 seconds
Time taken = 2.3 seconds
Let us assume that the car accelerates with 'a' which changes the velocity from 16m/s to 45m/s.
Acceleration = Final velocity - Initial velocity/Time taken
Acceleration = (45 m/s - 16 m/s)/2.3 seconds
Acceleration = 29/2.3 m/s²
Acceleration = a = 12.6 m/s²
Now let the distance traveled by car be x meters
Using the third equation of motion
v² - u² = 2ax
Put in the values, we get
45² - 16² = 2×12.6×(x)
2025 - 256 = 25.2x
25.2x = 1769
x = 1769/25.2
x = 70.2 meters
Therefore, the car travels a distance of 70.2 meters while accelerating from 16 m/s to 45m/s in 2.3 seconds.
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