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.

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 = (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!

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

To solve this problem, we need to integrate the given acceleration function to find the velocity function. Then we can use the velocity function to find the exact time when the particle's velocity is zero.

A. Finding the velocity function v(t):
Given acceleration function a(t) = e^(t/2) m/sec^2

To find the velocity function v(t), we need to integrate the acceleration function. Since the acceleration is given as a function of t, we integrate with respect to t.

∫ a(t) dt = ∫ e^(t/2) dt

To integrate e^(t/2), we can use the substitution method. Let u = t/2, then du = (1/2) dt.

Substituting back and simplifying, we have:

∫ e^(t/2) dt = 2 ∫ e^u du = 2e^u + C

Now substituting u = t/2 back into the expression, we get:

v(t) = 2e^(t/2) + C

To determine the constant C, we are given that the velocity at t=0 is -4 m/sec:

v(0) = 2e^(0/2) + C = 2 + C = -4

Solving for C, we find C = -6.

Therefore, the velocity function v(t) is:

v(t) = 2e^(t/2) - 6 m/sec

B. Finding the time when the velocity is zero:
To find the exact time when the velocity is 0 m/s, we can set v(t) equal to zero and solve for t:

0 = 2e^(t/2) - 6

Adding 6 to both sides:

6 = 2e^(t/2)

Dividing both sides by 2:

3 = e^(t/2)

To get rid of the exponential, we take the natural logarithm (ln) of both sides:

ln(3) = ln(e^(t/2))

Using the property of logarithms that ln(e^x) = x:

ln(3) = t/2

Multiplying both sides by 2:

t = 2ln(3)

Therefore, the exact time when the velocity is 0 m/s is t = 2ln(3) seconds.