Given the car moves in a straight line such that its velocity is defined by $v=0.3 (9t^2 + 2t)$ where 't' is in seconds

To find

• The position and acceleration when t = 3 sec. taking at$t=0s, x=0m$
• We know that$v = {dx \over dt} $and$a= {dv \over dt}$
• So$x = \int vdt$

$x = \int 0.3(9t^2+2t)dt=0.3(3t^3+t^2)+c$

• Given initial condition as t=0s, x=0m so c=0

$x(t=3s) = 0.3(81+9)=27m$

• Similarly

​ ​$a= \dfrac{\mathrm dv }{ \mathrm dt}=\dfrac{\mathrm d(0.3(9t^2+2t))}{\mathrm dt}=(5.4t+0.6 ) \ m/s^2$

$a(t=3s)=(5.4(3)+0.6 )m/s^2=16.8m/s^2$

Answer: The position and acceleration when t = 3 sec are 27m and 16.8m/s2 respectively.