Answer to Question #296666 in Molecular Physics | Thermodynamics for dandan

Question

The acceleration of a point is a= 30t m/s^2. When t= 0, s=50 m, and V= 15 m/s. a.)What is the position of the point at t=3 sec? b.) What is the velocity of the point at t= 3 sec? c.) What is the acceleration of the point at t= 3 sec? *

Answer 1

Solution;

$Given;$

$a=30t(m/s^2)$

At t=0,s=50m,V=15m/s

We know;

$a(t)=\frac{d^2s}{dt^2}=30t$

Integrate to obtain velocity;

$V(t)=\frac{ds}{dt}=\int30tdt=\frac{30t^2}{2}+c$

Apply the given condition;

$V(0)=15=15(0)+c$

Hence; c=15, and;

$V(t)=(15t^2+15)m/s$

Integrate the above equation to obtain position;

$s=\int15t^2+15=\frac{15t^3}{3}+15t+c$

Apply the given condition;

$s(0)=50=5(0)+15(0)+c$

Hence,c=50 and;

$s(t)=(5t^3+15t+50)m$

(a)

$s(3)=5(3^3)+15(3)+50=230m$

(b)

$V(3)=15(3^2)+15=150m/s$

(c)

$a(3)=30(3)=90m/s^2$