Question

At time t, the distance x cm of a particle moving in a horizontal line is given by x = 4t sq. + 2t. Find velocity and acceleration when t = 0.5 s.

Answer 1

Here,

X = 4$t^{2}$ + 2t

And therefore,

Velocity a time t = v = $\frac{dx}{dt}$ = 8t + 2.    

Acceleration at time t = a = $\frac{dv}{dt}$  =  8  

When t = 0.5 ,

Velocity = 8t + 2

       = 8(0.5) +2  

       = 6 cm/s

And ,

Acceleration= (8) = 8 cm/ $sec^{2}$

Answer 2

v=dx/dt

v=d(4t2+2t)/dt

v=8t+2

v=6

a=dv/dt

a=d(8t+2)/dt

a=8