A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.
1 answer:
Answer:
See the attachment below for the graphics in part (a)
The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.
This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.
Explanation:
The full solution can be found in the attachment below.
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The third equation of free fall can be applied to determine the acceleration. So that Paola's acceleration during the flight is 39.80 m/.
Acceleration is a quantity that has a direct relationship with velocity and also inversely proportional to the time taken. It is a vector quantity.
To determine Paola's acceleration, the third equation of free fall is appropriate.
i.e =
± 2as
where: V is the final velocity, U is the initial velocity, a is the acceleration, and s is the distance covered.
From the given question, s = 20.1 cm (0.201 m), U = 4.0 m/s, V = 0.
So that since Poala flies against gravity, then we have:
=
- 2as
0 = - 2(a x 0.201)
= 16 - 0.402a
0.402a = 16
a =
= 39.801
a = 39.80 m/
Therefore Paola's acceleration is 39.80 m/.