Question

A) If a particle's position is given by x = 4 -11t + 3t² (where t is in seconds and x is in meters), what is its velocity at t = 1.0 s?

B) Is it moving in the positive or negative direction of x just then?

C) What is its speed just then and is the speed increasing or decreasing just then?

D) At what time is the velocity equal to zero and what is the position at that time?

Answer 1

x = 4 -11t + 3t²

A) velocity is the rate of change of displacement with time

v = dx/dt

v = 6t - 11

at t = 1sec

v = 6x1 -11 = - 5m/s

B) it is moving in the negative x direction

c) Its speed is 5m/s and it is decreasing

D) if its velocity is zero then

6t - 11 = 0

hence t = 1.833 sec

the position will be

x = 4 -11t + 3t²

x = 4 -11(1.833) + 3(1.833)²

x = - 6.0833 m