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

Answer 2

Acceleration is the rate of change of velocity over time.

The position and velocity of the particle at t = 3s are 52m/s and 116m, respectively.

The given parameters are:

$\mathbf{a =20ms^{-2}}$ -- acceleration

$\mathbf{u =-8ms^{-1}}$ --- initial velocity

$\mathbf{s_1 =50m}$ --- the starting distance

The velocity at t = 3s is calculated using:

$\mathbf{v =u + at}$

So, we have:

$\mathbf{v =-8 + 20 \times 3}$

$\mathbf{v =-8 + 60}$

$\mathbf{v =52}$

The position at t = 3 is calculated using:

$\mathbf{s = s_1 + ut +\frac{1}{2}at^2}$

So, we have:

$\mathbf{s = 50 - 8 \times 3 +\frac{1}{2} \times 20 \times 3^2}$

$\mathbf{s = 50 - 24 +90 }$

$\mathbf{s = 116}$

Hence, the position and velocity of the particle at t = 3s are 52m/s and 116m, respectively.

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