Answer:
Vo = 4.5 [m/s]
Explanation:
In order to solve this problem, we must use the following equation of kinematics.
where:
Vf = final velocity = 12 [m/s]
Vo = initial velocity [m/s]
a = acceleration = 1.5 [m/s²]
t = time = 5 [s]
Now replacing:
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
Answer:
Explanation:
Given
coefficient of kinetic friction
mass of locomotive=180,000 kg
v=10 m/s
deceleration provided by friction
Time taken to stop completely is
v=u+at
where v=final velocity
u=initial velocity
a=acceleration
t=510.20 s
Distance traveled during this point
s=2551.02 m
Answer:
Please see list of answers below
Explanation:
If the FORCE applied to the car doubles, then following the equation F = m * a , the acceleration will DOUBLE as well.
The mass of the car stays the SAME.
And the velocity of the car will INCREASE with time based on the formula of kinematics of an accelerated object:
v(t) = vi + a * t
(Hint: the time<span> to rise to the </span>peak<span>is one-half the </span>total hang-time<span>.).</span>