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.