a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?
1 answer:
<u>Answer:</u>
The acceleration of the car is
<u>Explanation:</u>
In the question it is given that car initially heads north with a velocity . It then accelerates for
and in the end its velocity is
.
initial velocity
time
final velocity
The equation of acceleration is
The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.
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Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F =
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2
(d-r) ² = r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 - = 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r =
r = [2d ± 2d ] / 2a
r = (1 ± √ (1.65 10⁻³)) =
(1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ = 1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m