Question

The acceleration of a point is given. a = 20 t m/s2 When t=0, s = 50 m and v = -8 m/s. What are the position and velocity of the particle at t = 3 s?

Answer 1

Answer:

v=82 m/s

s=116m

Explanation:

a=20t

$\frac{\mathrm{d} v}{\mathrm{d} t}=20t\\\int\limits dv =\int(20t) dt\\v={10}t^2+c$

using condition given at t=0

$-8=10\times 0^2 +c$

c=-8

now equation becomes

v=10t²-8

v at t= 3s  v=82 m/s

again

$\frac{\mathrm{d} s}{\mathrm{d} t}=10t^2-8$

$ds=(10t^2-8)dt\\\int\limits ds =\int(10t^2-8) dt\\s=\frac{10}{3} t^2-8t+b$

now using condition given s=50 at t=0

b=50

now equation becomes

$s=\frac{10}{3}t^3-8t+50$

calculating s at t=3s

s=116m