Question

The displacement (in meters) of a particle moving in a straight line is given by s=6t^3+1t+4 where t is measured in seconds.

Find the velocity of the particle at the times given.
Velocity at time t=1, t=2, t=3, and t=a

Answer 1

Answer:
Velocities

t = 1 : 11 m/s

t = 2 : 27m/s

t = 3: 56.33 m/s

t = a: 6a + 1 + 4/a m/s

Explanation:

s(t) = 6t³ + 1t + 4 where s = displacement in meters and t is time in seconds

s(t) = v(t). t where v is the velocity in m/s and t is time in seconds
So v = s/t

At time t = 1
s(1) = 6 x 1³ + 1 x 1 + 4 = 6 + 1 + 4 = 11 meters
v(1) = 11/1 = 11 m/s

At time t = 2
s(2) = 6 x 2³ + 1 x 2 + 4 = 6 x 8 + 2 + 4 = 48+ 2 + 4 = 54 m
v(2) = 54 /2 = 27m/s

At time t = 3
s(3) = 6 x 3³ + 1 x 3 + 5 = 6 x 27 + 3 + 4 = 169 m
v(3) = 169/3 = 56.33 m/s

At time t = a
s(a) = 6a³ + 1a + 4 = 6a³ + a + 4
v(a) = s(a)/a = (6a³ + a + 4)/a = 6a + 1 + 4/a m/s