Question

If u(t) = 6t 1.9 m/s and s = 2 m when t 0, find expressions for a(t) and s(t). Calculate values for s(t), v(t), and a(t) for times t = 0, 1, 2, and 3 seconds and tabulate the results.

Answer 1

Answer:

t   in s                   0                    1                    2                    3

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s  in m                  2                  4.068          17.44              55.05

a  in m/s²             0                    11.4            21.27             30.64

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Explanation:

Given:

u(t) = $6t^{1.9}$

s = 2 m at t = 0

Now,

u(t) = $\frac{\textup{ds}}{\textup{dt}}$

thus,

$\frac{\textup{ds}}{\textup{dt}}=6t^{1.9}$

or

$ds=6t^{1.9}dt$

on integrating, we get

$\int \, ds=\int{6t^{1.9}} \, dt$

or

s = $\frac{6\ t^{1.9+1}}{1.9+1}+c$

here c is the integration constant

s = $\frac{6\ t^{2.9}}{2.9}+c$

now, at t = 0, s = 2 m

thus,

2 = $\frac{6\times0^{2.9}}{2.9}+c$

or

c = 2

hence,

the expression is

s = $\frac{6\timest^{2.9}}{2.9}+2$

also,

a(t) = $\frac{\textup{du}}{\textup{dt}}$

or

a(t) =  $\frac{\textup{d(6t^{1.9})}}{\textup{dt}}$

or

a(t) =   $6\times1.9\times t^(1.9-1)$

or

a(t) = $11.4\times t^(0.9)$

now,

at t = 0

s =  $\frac{6\times0^{2.9}}{2.9}+2$

or

s = 2 m

and,

a = $11.4\times 0^(0.9)$

or

a = 0 m/s²

at t = 1

s =  $\frac{6\times1^{2.9}}{2.9}+2$

or

s = 4.068 m

and,

a = $11.4\times 1^(0.9)$

or

a = 11.4 m/s²

at t = 2

s =  $\frac{6\times2^{2.9}}{2.9}+2$

or

s = 17.44 m

and,

a = $11.4\times 2^(0.9)$

or

a = 21.27 m/s²

at t = 3

s =  $\frac{6\times3^{2.9}}{2.9}+2$

or

s = 55.05 m

and,

a = $11.4\times 3^(0.9)$

or

a = 30.64 m/s²

t   in s                   0                    1                    2                    3

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s  in m                  2                  4.068          17.44              55.05

a  in m/s²             0                    11.4            21.27             30.64

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