Acceleration occurs when there is a change in speed or direction. If it travels in a straight line, there is no speed or change in direction as it is constant througout, hence 0 acceleration.
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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?
Time t = ?
<span>When wave is moving from
y = 0 to y =12 cm</span>
By using the formula,
y = 15cos [(π/12) t)] =
0,
cos [(π/12) t)] = 0 =
cos (π/2), so,
(π/12)t = π/2,
t = (π/2) (12/π)
t = 12/2
<span>t = 6 sec</span>
<span>so 6 sec is the least amount of time required</span>
A net torque of magnitude is 600
Juice WRLD 999 ,........................
