postition= integral v dt
v= int a dt
find v first, you know the constant of integration from at t=2
Then integrate again to get position.
The acceleration of a particle at a time t moving along the x-axis is give by: a(t) = 4e^(2t). At the instant when t=0, the particle is at the point x=2, moving with velocity v(t)=-2. Find the position of the particle at t=1/2
if you could show me how to get that please
2 answers
a(t) = 4e^(2t)
so v(t) = 2e^(2t) + C (I integrated, since dv/dt = a)
when t=0, v=-2
-2 = 2e^0 + c
c = -4 and then
v)t) = 2e^(2t) - 4
since ds/dt = v
s = e^(2t) - 4t + k
when t=0, s=2 (I assume that is what you meant by x=2)
2 = e^0 - 4(0) + k
k = 1
then s(t) = e^(2t) - 4t + 1
s(1/2) = e^1 - 2 + 1
= e - 1
so v(t) = 2e^(2t) + C (I integrated, since dv/dt = a)
when t=0, v=-2
-2 = 2e^0 + c
c = -4 and then
v)t) = 2e^(2t) - 4
since ds/dt = v
s = e^(2t) - 4t + k
when t=0, s=2 (I assume that is what you meant by x=2)
2 = e^0 - 4(0) + k
k = 1
then s(t) = e^(2t) - 4t + 1
s(1/2) = e^1 - 2 + 1
= e - 1
1 answer