Give acceleration a=d^2s/dt^2,initial velocity, and initial position of a body moving on a coordinate line. Find the bodys position at time t.
a=e^t v(0)=20 s(0)=5
To find the body's position at time t, we need to integrate the acceleration function twice. Given the initial velocity and initial position, we can use these values to determine the constants of integration for each integration step.
Step 1: Integration to find velocity (v) as a function of time (t)
We're given that the initial velocity (v(0)) is 20, so we integrate the acceleration function a(t) = e^t with respect to time (t):
∫(a(t)) dt = v(t)
∫(e^t) dt = v(t)
v(t) = ∫(e^t) dt
Using the integral of e^t, we get:
v(t) = e^t + C1
where C1 is the constant of integration.
Given v(0) = 20, we can substitute the initial conditions to solve for C1:
v(0) = e^0 + C1
20 = 1 + C1
C1 = 20 - 1
C1 = 19
Thus, the velocity function is given by:
v(t) = e^t + 19
Step 2: Integration to find position (s) as a function of time (t)
We're given that the initial position (s(0)) is 5, so we integrate the velocity function v(t) = e^t + 19 with respect to time (t):
∫(v(t)) dt = s(t)
∫(e^t + 19) dt = s(t)
Using the integral of e^t, we get:
s(t) = ∫(e^t + 19) dt
s(t) = e^t + 19t + C2
where C2 is the constant of integration.
Given s(0) = 5, we can substitute the initial conditions to solve for C2:
s(0) = e^0 + (19 * 0) + C2
5 = 1 + C2
C2 = 5 - 1
C2 = 4
Thus, the position function is given by:
s(t) = e^t + 19t + 4
Hence, the body's position at time t is e^t + 19t + 4.