velocity of a particle- the displacement s (in meters) of a particle moving in a straight line is given by the equation of motion s=4t^3+6t+2, where t is measured in seconds. Find the velocity of the particle s at times t=a t=1 t=2 t=3
same as the one before
thanks
im stuck still
s = 4t^3 + 6t + 2
ds/dt = v = 12t^2 + 6
when t = 1, v = 12(1^2) + 6 = 18
when t = 2, v = ....
when t = 3 .......
when t = a , b = 12a^2 + 6
thank you
To find the velocity of the particle at different times (t = a, t = 1, t = 2, t = 3), we need to calculate the derivative of the displacement equation with respect to time (t).
The velocity (v) of a particle is the rate of change of its displacement (s) with respect to time (t). Mathematically, it can be expressed as:
v = ds/dt
Taking the derivative of the displacement equation s = 4t^3 + 6t + 2 with respect to t, we get:
v = d/dt(4t^3 + 6t + 2)
Differentiating each term separately:
v = d/dt(4t^3) + d/dt(6t) + d/dt(2)
Using the power rule of differentiation, we can find the derivatives of each term:
v = 12t^2 + 6 + 0
Simplifying, we get:
v = 12t^2 + 6
Now, we can substitute specific values of t into the equation to find the velocity at those times.
a) For t = a:
v = 12a^2 + 6
b) For t = 1:
v = 12(1)^2 + 6
= 12 + 6
= 18
c) For t = 2:
v = 12(2)^2 + 6
= 12(4) + 6
= 48 + 6
= 54
d) For t = 3:
v = 12(3)^2 + 6
= 12(9) + 6
= 108 + 6
= 114
Therefore, the velocity of the particle at times t = a, t = 1, t = 2, and t = 3 are given by the expressions 12a^2 + 6, 18, 54, and 114, respectively.