A rocket takes off vertically at time t=0, and during the first 10 s of flight its acceleration in m/s^2 is given by a(t)=10t-t^2 find the height reached by the rocket after 10s.
hn bgm.
njk.v
To find the height reached by the rocket after 10 seconds, we need to integrate the function for acceleration with respect to time.
Given that the acceleration is given by a(t) = 10t - t^2, we can integrate it to find the velocity function v(t) and then integrate again to find the position function s(t).
1. Finding the velocity function:
To find the velocity, we integrate the acceleration function:
v(t) = ∫(a(t)) dt
= ∫(10t - t^2) dt
Using the power rule of integration, we get:
v(t) = 5t^2 - (1/3)t^3 + C
where C is the constant of integration.
2. Finding the position function:
To find the position, we integrate the velocity function:
s(t) = ∫(v(t)) dt
= ∫(5t^2 - (1/3)t^3 + C) dt
Using the power rule of integration again, we get:
s(t) = (5/3)t^3 - (1/12)t^4 + Ct + D
where C and D are constants of integration.
Now, let's apply the initial conditions. At t = 0, the rocket starts from rest, so v(0) = 0. Using this condition, we can solve for the constant of integration C:
v(0) = 5(0)^2 - (1/3)(0)^3 + C
0 = 0 + 0 + C
C = 0
Therefore, the velocity function becomes:
v(t) = 5t^2 - (1/3)t^3
Now, let's apply the second initial condition. After 10 seconds, we want to find the height reached by the rocket. The position function s(t) will give us that information:
s(10) = (5/3)(10)^3 - (1/12)(10)^4 + 0 * 10 + D
= 5000/3 - 2500/3 + D
= 2500/3 + D
Since we want the height and not the displacement, we need to find the relative height above the starting point. This is given by s(10) - s(0):
Relative height = s(10) - s(0)
= (2500/3 + D) - (0 + D)
= 2500/3
Therefore, after 10 seconds, the rocket reaches a height of 2500/3 units.