Part A:
Well, if you integrate acceleration, you get velocity, so this part is simple:
∫[e^(t/2)]dt = (1/2)e^(t/2)+C
At t=0, the velocity is -4 m/sec, so we need to solve for the constant C to complete the velocity function v(t):
(1/2)e^(0/2) + C = -4
1/2 + C = -4
C = -9/2
Thus, the function is v(t) = (1/2)e^(t/2) - 9/2
Part B:
By setting v(t)=0, we can find the exact time that that occurs:
v(t) = (1/2)e^(t/2) - 9/2
0 = (1/2)e^(t/2) - 9/2
9/2 = (1/2)e^(t/2)
9 = e^(t/2)
ln(9) = t/2
2ln(9) = t
Thus, the exact time when the velocity of the particle is 0 m/s is 2ln(9) seconds.
Hope these answers and explanations helped!
A particle moves on a coordinate line with an acceleration at time t seconds of e^t/2 m/sec^2. At t=0 the particle is at the origin, and its velocity is -4 m/sec.
A. Find a function v(t) that gives the velocity of the particle at time t.
B. At what (exact) time is the velocity of the particle 0 m/s?
I'm having trouble with this problem, and would really appreciate some help.
3 answers
I will read that as you are given:
a = (1/2) e^t m/s^2
v = 1/2 e^t + c
when t = 0, v(0) = -4 m/s
-4 = 1/2 (1) + c
c = -9/2
v(t) = 1/2 e^t - 9/2
b) when is v(t) = 0 ?
1/2 e^t - 9/2 = 0
e^t = 9
t = ln9 = 2.2
at 2.2 s, v = 0
If you meant
a = e^(t/2) , it would change everything
v = 2 e^(t/2) + k
-4 = 2(1) + k
k = -6
etc
a = (1/2) e^t m/s^2
v = 1/2 e^t + c
when t = 0, v(0) = -4 m/s
-4 = 1/2 (1) + c
c = -9/2
v(t) = 1/2 e^t - 9/2
b) when is v(t) = 0 ?
1/2 e^t - 9/2 = 0
e^t = 9
t = ln9 = 2.2
at 2.2 s, v = 0
If you meant
a = e^(t/2) , it would change everything
v = 2 e^(t/2) + k
-4 = 2(1) + k
k = -6
etc
Let me redo this problem (I accidentally did differentiation instead of integration lol):
Part A:
Well, if you integrate acceleration, you get velocity, so this part is simple:
∫[e^(t/2)]dt = 2e^(t/2)+C
At t=0, the velocity is -4 m/sec, so we need to solve for the constant C to complete the velocity function v(t):
2e^(0/2) + C = -4
2 + C = -4
C = -6
Thus, the function is v(t) = (1/2)e^(t/2) - 6
Part B:
By setting v(t)=0, we can find the exact time that that occurs:
v(t) = 2e^(t/2) - 6
0 = 2e^(t/2) - 6
6 = 2e^(t/2)
3 = e^(t/2)
ln(3) = t/2
2ln(3) = t
Thus, the exact time when the velocity of the particle is 0 m/s is 2ln(3) seconds.
Hope these answers and explanations helped!
Part A:
Well, if you integrate acceleration, you get velocity, so this part is simple:
∫[e^(t/2)]dt = 2e^(t/2)+C
At t=0, the velocity is -4 m/sec, so we need to solve for the constant C to complete the velocity function v(t):
2e^(0/2) + C = -4
2 + C = -4
C = -6
Thus, the function is v(t) = (1/2)e^(t/2) - 6
Part B:
By setting v(t)=0, we can find the exact time that that occurs:
v(t) = 2e^(t/2) - 6
0 = 2e^(t/2) - 6
6 = 2e^(t/2)
3 = e^(t/2)
ln(3) = t/2
2ln(3) = t
Thus, the exact time when the velocity of the particle is 0 m/s is 2ln(3) seconds.
Hope these answers and explanations helped!
5 answers