A car is traveling at 19m/s. It slows to a stop at a constant rate over 4.2 seconds. How far did the car travel during the 4.2se
1 answer:
The distance covered by the car is 39.9 m
Explanation:
Since the car is slowing at constant rate, it means that its acceleration is constant, therefore we can solve the problem by using the following suvat equation:
where
s is the distance covered
u is the initial velocity
v is the final velocity
t is the time elapsed
For the car in this problem, we have
u = 19 m/s (initial velocity)
v = 0 (final velocity)
t = 4.2 s (time elapsed)
Therefore, the distance covered is
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The gravitational force between the two objects A) It increases.
Explanation:
The gravitational force between two objects is given by:
(1)
where
G is the gravitational constant
are the masses of the two objects
r is the separation between the objects
In this problem, object A and object B are initially at a distance of
r = 100 m
And at that distance, the force between them is
F
Later, object A gains some mass. We notice from eq.(1) that the gravitational force is directly proportional to the mass: therefore, if the mass of either of the two objects increases, then the gravitational force between them also increases. Therefore, the new force will be larger than the original force:
F' > F
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