Hi,

Can anyone offer any advice on solving the following,

motion of two stage rocket is stated by it height y in metres above the ground since t seconds since launch.

Flies vertically Ist stage burns for 5 seconds and cuts out. 2nd stage rocket is under gravity alone.

modelled by the equations

Stage 1 height y = 10t2

time 0< t <5

stage 2 height y = -5t2 + 150t - 375

time t >5

Can anyone offer some help... I need to work out speed velocity etc,

Thank you

during the first 5 seconds the height is given by 10 t^2

So it reaches an altitude of 10(25) = 250 meters before dropping the first stage.
At that time, 5 seconds, the velocity is given by dy/dt = 20 t = 100 m/s

Then stage two continues. It shows no sign of being a rocket since it seems to be operating only under gravity with an initial height of 250 meters at t = 5 and an initial velocity of 100 m/s at t = 5
Its velocity is determined by
dy/dt = -10 t + 150
at the maximum height, that velocity will be zero
0 = -10 t + 150
or t = 15 seconds at peak
then the max height will be
y = -5 (225) + 150 (15) - 375
y = -1125+2250-375
y = 750 meters at peak.
Then when will it hit the ground?
0 = -5 t^2 + 150 t -375
t^2 -30 t + 75 = 0
t=(1/2) [30 +/-sqrt(900-300)]
t = (1/2)[30 +/- 24.5]
Well, the minus sign will not do because the thing is still headed with stage 1 up so use the + sign
t = (1/2)(54.5) = 27.25 seconds to crash

If you mean

y = 10t^2 for 0<t<5 s and
y = -5t^2 +150 t -375 for t>5 (until t = 25 s, when y = 0), then
the velocity (and speed) are
20 t for t<5 and
-10t + 150 for 5<t<25

The acceleration is 20 during t<5s and -10 during 5<t<25s

My answers were obtained by differentiation. I assume you are studying calculus.

Thank you for your help

My statement that y=0 at t=25 s is incorrect. Damon derived the correct value

Thank you ..

Thank you have a good weekend !!

To solve for speed and velocity, you first need to understand the basic concepts of calculus. This problem involves calculus because you want to find the derivative of the height equation to obtain the velocity, and the second derivative to obtain the acceleration.

Let's start by finding the velocity. Velocity is the derivative of the position function with respect to time. In this case, we have two different equations for two different time intervals.

For the first stage (0 < t < 5), we have the equation:
y = 10t^2

To find the velocity, we need to take the derivative of this equation with respect to time. So we have:
dy/dt = d(10t^2)/dt

To take the derivative, you can apply the power rule of differentiation, which states that the derivative of t^n is n*t^(n-1). Applying this rule, we get:
dy/dt = 20t

So, the velocity during the first stage is given by:
v1(t) = 20t

For the second stage (t > 5), we have the equation:
y = -5t^2 + 150t - 375

Again, we need to take the derivative of this equation with respect to time:
dy/dt = d(-5t^2 + 150t - 375)/dt

Taking the derivative of each term, we get:
dy/dt = -10t + 150

So, the velocity during the second stage is given by:
v2(t) = -10t + 150

To find the speed, you need to take the absolute value of the velocity since speed is always a positive value. So, the speed during the first stage is:
s1(t) = |v1(t)| = 20t

And the speed during the second stage is:
s2(t) = |v2(t)| = |-10t + 150|

To find acceleration, you need to take the derivative of the velocity functions obtained above. The acceleration is the second derivative of the position function.

For the first stage:
a1(t) = dv1(t)/dt = d(20t)/dt = 20

Throughout the entire second stage, the rocket is only under the influence of gravity. So, the acceleration is just the acceleration due to gravity, which is approximately 9.8 m/s^2.

a2(t) = 9.8

I hope this helps! Let me know if you have any further questions.