I. Positive acceleration increases velocity. Negative acceleration decreases velocity. runner A sped up until the finish line and then slowed to a stop.
ii. Zero a acceleration implies a constant, unchanging velocity not a zero velocity. runner B achieved some velocity prior to 8s and is moving and must slow down to reach a stop.
iii. None. No aspects of this reasoning are correct. Everything she says is wrong. See iv for what/why.
iv. The sign on acceleration denotes the direction of *change in velocity* not change in direction. The sign on velocity can denote change in direction but only “forward” or “reverse” along a particular path. Cardinal direction is not indicated, generally, by the sign on velocity. It may correspond to North/South situationally but it is not an built-in feature of velocity and its sign. For example, if you are traveling with positive velocity and turn left to continue your journey you still have a positive velocity in the new direction. In fact, if you turn left again, traveling in the opposite direction as the one you started with your velocity would still be positive… in the new direction. The velocity relative to original direction could be said to be negative but that would be a confusing way to describe a journey. Maybe if you stopped the vehicle and moved in reverse, you could meaningfully say velocity was negative.
Answer:
Explanation:
Given that a centripetal force is a form of force that gives rise or causes a body to move in a curved path.
Hence;
1. When a car is being driven around a track, it is the FORCE OF FRICTION that is acting upon the turned wheels of the vehicle, which transforms into the centripetal force required for circular motion.
2. When a ball being is swung on the end of a string, TENSION FORCE acts upon the ball, which transforms the centripetal force required for circular motion.
3. When the moon is orbiting the earth, it is the FORCE OF GRAVITY acting upon the moon, which transforms the centripetal force required for circular motion.
4. A rotating wheel on the other hand has NO centripetal force because centripetal force is pull towards the center of a motion. However the speed of the object is tangent to the circle, while the direction of the force is also perpendicular to the direction of the rotating wheel.
This means acceleration a is constant.
Let
a) vo be the initial speed, at t=0
b) v be the final speed after time t
c) d distance travelled in time t
Then we have:
a) v=vo+a×t
b) v²=vo²+2×a×d (Galilei's equation)
c) d=vo×t+a×t²/2
d) average speed vm=(vo+v)/2
Yp(t) = A1 t^2 + A0 t + B0 t e(4t)
=> y ' = 2A1t + A0 + B0 [e^(4t) +4 te^(4t) ]
y ' = 2A1t + A0 + B0e^(4t) + 4B0 te^(4t)
=> y '' = 2A1 + 4B0e(4t) + 4B0 [ e^(4t) + 4te^(4t)
y '' = 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t)
Now substitute the values of y ' and y '' in the differential equation:
<span>y′′+αy′+βy=t+e^(4t)
</span> 2A1 + 4B0e^(4t) + 4B0e^(4t) + 16B0te^(4t) + α{2A1t + A0 + B0e^(4t) + 4B0 te^(4t) } + β{A1 t^2 + A0 t + B0 t e(4t)} = t + e^(4t)
Next, we equate coefficients
1) Constant terms of the left side = constant terms of the right side:
2A1+ 2αA0 = 0 ..... eq (1)
2) Coefficients of e^(4t) on both sides
8B0 + αB0 = 1 => B0 (8 + α) = 1 .... eq (2)
3) Coefficients on t
2αA1 + βA0 = 1 .... eq (3)
4) Coefficients on t^2
βA1 = 0 ....eq (4)
given that A1 ≠ 0 => β =0
5) terms on te^(4t)
16B0 + 4αB0 + βB0 = 0 => B0 (16 + 4α + β) = 0 ... eq (5)
Given that B0 ≠ 0 => 16 + 4α + β = 0
Use the value of β = 0 found previously
16 + 4α = 0 => α = - 16 / 4 = - 4.
Answer: α = - 4 and β = 0
If the object sinks, then it must be heavier than the weight of the water
it displaces ... heavier than the buoyant force acting on it.
If the buoyant force were equal or greater than the object's weight, then
the object would rise to the surface in water.
