Given the y coordinate of a particle is given by $y=6t^3-5t$. 

$a_x=14t m/sec^2$ & $ v_x=4 m/sec$ at t=0,

To find the velocity & acceleration of particle when t=1 second.

Solution:

We know that,

• Velocity in y direction$v_y={dy\over dt}={d(6t^3-5t)\over dt}=(18t^2-5)m/s$

Acceleration in y direction $a_y={dv\over dt}={d^2y\over dt}={d(18t^2-5)\over dt}=36t$

• Velocity in x direction$u_x=v_x+a_xt=4+14t^2$

Acceleration in x direction $a_x=14tm/sec^2$

• When t=1s

Velocity in y direction $v_y(t=1s)=18(1^2)-5=13m/s$

Velocity in x direction $v_x(t=1s)=4+14(1^2)=18m/s$

• Acceleration in y direction$a_y(t=1s)=36(1)=36m/s^2$
• Acceleration in x direction$a_x(t=1s)=14(1)=14m/s^2$
• So the velocity of particle at t=1s is

$V=(18i+13j)m/s$

• So the acceleration of particle at t=1s is

$A=(14i+36j)m/s^2$