Sure. The acceleration may be decreasing, but as long as it stays
in the same direction as the velocity, the velocity increases.
I think you meant to ask whether the body can have increasing velocity
with negative acceleration. That answer isn't simple either.
If the body's velocity is in the positive direction, then positive acceleration
means speeding up, and negative acceleration means slowing down.
BUT ... If the body's velocity is in the negative direction, then positive
acceleration means slowing down, and negative acceleration means
speeding up.
I know that's confusing.
-- Take a piece of scratch paper, write a 'plus' sign at one edge and
a 'minus' sign at the other edge. Those are the definitions of which
direction is positive and which direction is negative.
-- Then sketch some cars ... one traveling in the positive direction, and
one driving in the negative direction. Those are the directions of the
velocities.
-- Now, one car at a time:
. . . . . first push on the back of the car, in the direction it's moving;.
. . . . . then push on the front of the car, against its motion.
Each push causes the car to accelerate in the direction of the push.
When you see it on paper, all the positive and negative velocities
and accelerations will come clear for you.
No...........................................................................Work for them urself
The problem states that the distance travelled (d) is
directly proportional to the square of time (t^2), therefore we can write this in
the form of:
d = k t^2
where k is the constant of proportionality in furlongs /
s^2
<span>Using the 1st condition where d = 2 furlongs, t
= 2 s, we calculate for the value of k:</span>
2 = k (2)^2
k = 2 / 4
k = 0.5 furlongs / s^2
The equation becomes:
d = 0.5 t^2
Now solving for d when t = 4:
d = 0.5 (4)^2
d = 0.5 * 16
<span>d = 8 furlongs</span>
<span>
</span>
<span>It traveled 8 furlongs for the first 4.0 seconds.</span>
Answer:
r=6.05km/hr
z=59.1 degree to the horizontal
Explanation:
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place
can be solved using pythagoras theorem
r^2=o^2+a^2
r^2=5.2^2+3.1^2
r^2=36.65
r=6.1km/hr is te birds velocity relative to the ground
tanz=5.2/3.1
z=tan^-1(5,2/3.1)
z=59.1 degree to the horizontal
