a particle moves according to the law a=1/t √t. if s=8 when t=1. and s=16 when t=4. find v when t=9.

do we have a = 1/(t√t) or is it a = (1/t)√t ??

a = 1/(t√t)

ok, then

a = t^(-3/2)
v = -2 t^(-1/2) + c

s = -4 t^(1/2) + ct + k

case1: s=8, when t = 1
8 = -4 + c + k
12 = c + k

case2: s = 16, when t = 4
16 = -4(2) + 4c + k
24 = 4c + k

subtract them:
12 = 3c
c = 4

back in 12 = c+k
12 = 4+k
k = 8

then v = -2/√t + 4
when t = 9
v = -2/3 + 4 = 10/3

Thank you very, very much. I'm really grateful.

welcome,

hope you checked my arithmetic

To solve this problem, we need to find the velocity (v) when t=9. To do this, we can follow these steps:

Step 1: Find the acceleration (a) at t=1 and t=4:
At t=1, we have s=8. Since s=∫v dt (where ∫ represents integration), we can find v by differentiating s with respect to t:
v = ds/dt. Differentiating s=8 with respect to t gives v = 0.

Similarly, at t=4, we have s=16. Differentiating s=16 with respect to t gives v = 0.

Since v=0 at both t=1 and t=4, the particle is momentarily at rest at those instants.

Step 2: Find the acceleration (a) at t=9:
Given the acceleration function a = 1/t √t, we can substitute t=9 into the equation to find a:
a = 1/9 √9 = 1/9*3 = 1/27.

Step 3: Find the velocity (v) when t=9:
To find the velocity (v) when t=9, we can integrate the acceleration function with respect to time:
v = ∫a dt.

Integrating a=1/27 with respect to t gives:
v = ∫(1/27) dt
v = (1/27) ∫dt
v = (1/27) t + C,

where C is the constant of integration.

Since we know that v=0 when t=1, we can substitute those values into the equation to solve for C:
0 = (1/27) (1) + C
0 = 1/27 + C
C = -1/27.

Therefore, the equation for v when t=9 becomes:
v = (1/27) t - 1/27.

Substituting t=9 into this equation gives:
v = (1/27) (9) - 1/27
v = 9/27 - 1/27
v = 8/27.

So, the velocity when t=9 is 8/27.