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

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<6s, spacecraft A’s velocity vA and spacecraft B’s velocity vB as functions of t are given by the equations vA(t)=t^2−5t+20 andvB(t)=t^2+3t+10, respectively, where both velocities are in units of meters per second. At t=6s, 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 BB?

(b) Each spacecraft has thrusters for speeding up and reverse thrusters for slowing down. At t=0, one of the spacecrafts is slowing down and one is speeding up.

i. Which spacecraft is slowing down?

ii. Calculate the time interval during which the spacecraft selected in part (b)(i) is slowing down.

(c) Calculate the value of t at which the two spacecrafts are traveling at the same speed.

(d) Is the acceleration of spacecraft A increasing, decreasing, or constant?

Explanation:

Given that

for 0< t<6s

a) at t = 0

speed of A is greter than B at t = 0

b)

The speed of A is less by 5m/s²

i) A is slowing down

ii)

hence, A is slowing down from 0 to 2.5sec

c)

Hence, both are travelling at same speed at t = 1.25sec

d)

Hence the acceleration of space craft A is increasing

heats surface water up

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft