Question

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

Answer 1

Answer:

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

$V_A(t) =t^2-5t+20\\\\V_B(t)=t^2+3t+10$

for 0< t<6s

a) at t = 0

$V_A(0) = 20m/s\\\\V_B(0) = 10m/s$

speed of A is greter than B at t = 0

b)

$\frac{dV_A(t)}{dt} = 2t-5\\\\\frac{dV_B(t)}{dt} = 2t+3$

The speed of A is less by 5m/s²

i) A is slowing down

ii) $\frac{dV_B(t)}{dt} =2t-5\\\\t =5/2 = 2.5sec$

hence, A is slowing down from 0 to 2.5sec

c)

$V_A(t) = V_B(t)\\t^2-5t+20=t^2+3t+10\\\\10=8t\\t=1.25sec$

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

d)

$\frac{d^2V_B(t)}{dt^2} =\frac{d(2t-5)}{dt} =2>0$

Hence the acceleration of space craft A is increasing