A motion of a particle is given by the relationship a=6t. If s=8m and v=4m/s when t=1sec determine:

a) v-s relationship
b) s-t relationship

s is distance
V is velocity
T is time

a) v-s relationship is v = ds/dt. Did you mean to say v-t relationship? If so that is:

v = ∫ a dt
v = ∫ (6t) dt
v = 3t^2 + c
4 = 3(1)^2 + c
c = 1
v = 3t^2 + 1

b) s = ∫ v dt
s = ∫ (3t^2 + 1) dt
Continue similarly to a) above.

To solve this problem, we need to use the equations of motion.

a) To find the v-s relationship:
We know that acceleration (a) is the derivative of velocity (v) with respect to time (t). In other words, a = dv/dt.

Integrating both sides with respect to time, we get:

∫ a dt = ∫ dv

Integrating, we have:

∫ 6t dt = ∫ dv

To find the integral of 6t, we integrate term by term:

3t^2 + C₁ = v

Based on the given information, when t = 1 sec, v = 4 m/s:

3(1)^2 + C₁ = 4

Simplifying the equation, we find:

C₁ = 1

Therefore, the v-s relationship is:

v = 3t^2 + 1

b) To find the s-t relationship:
Velocity (v) is the derivative of position (s) with respect to time (t). In other words, v = ds/dt.

Integrating both sides with respect to time, we get:

∫ v dt = ∫ ds

Integrating, we have:

∫ (3t^2 + 1) dt = ∫ ds

To find the integral of 3t^2 + 1, we integrate term by term:

t^3 + t + C₂ = s

Based on the given information, when t = 1 sec, s = 8 m:

(1)^3 + (1) + C₂ = 8

Simplifying the equation, we find:

C₂ = 6

Therefore, the s-t relationship is:

s = t^3 + t + 6

So, the v-s relationship is v = 3t^2 + 1, and the s-t relationship is s = t^3 + t + 6.