Question

4. (a) If a particle's position is given by x(t) = 4-12t+ 3t^2(where t is in seconds and x
is in meters), what is its velocity at t = 1s? (b) Is it moving in the positive or negative
direction of x just then? (c) What is its speed just then? (d) Is the speed increasing
or decreasing just then?​

Answer 1

Answer:

a] -6 m/s     b]Negative        c] 6 m/s         d]Speed will continuously increase

Explanation:

a] As the equation of displacement is given, we will integrate this equation to get the equation of velocity-

     Equation of velocity:- v=dx/dt= 4-12t+3t^2

                                         v= -12+6t , Hence velocity at t=1 sec will be -6m/s[by putting t=1 in equation]

b] As the value of velocity is in negative, hence the particle is moving in negative direction.

c]Speed is just the magnitude of velocity, hence speed will be 6 m/s.

d]Speed will increase uniformly with time as the time increases.