Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
Answer:
The acceleration is a = 2.75 [m/s^2]
Explanation:
In order to solve this problem we must use kinematics equations.
where:
Vf = final velocity = 13 [m/s]
Vi = initial velocity = 2 [m/s]
a = acceleration [m/s^2]
t = time = 4 [s]
Now replacing:
13 = 2 + (4*a)
(13 - 2) = 4*a
a = 2.75 [m/s^2]
Matter, substance. Material howya call it.
Beginning when the bottom of the object first touches the water,
and as it descends and more and more of it goes under, the
buoyant force on it increases during that time.
As soon as the object is completely underwater, it doesn't matter
how deep under it is, the buoyant force on it remains the same.
That is not a question but not all scientific theories have stood the test of time
