While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. The time taken by the car will be 4 seconds. the correct answer is option(c).
When acceleration is constant, the rate of change in velocity is also constant. In the absence of any acceleration, velocity remains constant. When acceleration is positive, velocity becomes more significant.
Let a denote acceleration, u denote initial velocity, v denote final velocity, and t denote time.
The equation of motion is stated as,
v = u + at
v² = u² + 2as
A car travels across a distance of 60.0 m while accelerating constantly from 12.0 m/s to 18.0 m/s.
Then the time taken by the car will be,
u = 12 m/s
v = 18 m/s
s = 60 m
Put these in the equation v² = u² + 2as.
18² = 12² + 2 x a x 60
a = 1.5
Then the time will be
18 = 12 + 1.5t
1.5t = 6
t = 4 seconds
Hence, the time taken is 4 s.
The complete question is:
While accelerating at a constant rate from 12.0 m/s to 18.0 m/s, a car moves over a distance of 60.0 m. How much time does it take?
1.00 s
2.50 s
4.00 s
4.50 s
To know more about acceleration refer to: brainly.com/question/12550364
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