Question

please help me with this as fast as you can i have test and please show work

The position of an object moving in a straight line is given by x = -1.5 + 3.0t - 14t^2 + 5t^3, where x is in meters and t is in seconds. (a) What is the position of the object t = 1, 2, 3 and 4 s ? (b) What is the object's displacement between t = 2 and t = 4 s ? (c) What is the average velocity for the time interval from t = 2 and t = 4 s? (d) What is the instantaneous velocity at t = 1, 2, 3, and 4 s? (e) What is the average acceleration for the time interval from t = 2 to t = 4?

Answer 1

Answer 2

(a) Positions of the object at t = 1, 2, 3,4 seconds are taken as x_1, x₂ , x₃ ,x₄

$x_{1} = -1.5 + 3.0\times 1-14\times 1^2+5\times 1^3= - 7.5 m$

$x_{2} = -1.5 + 3.0\times 2-14\times 2^2+5\times 2^3= -11.5 m$

$x_{3} = -1.5 + 3.0\times 3-14\times 3^2+5\times 3^3= 16.5 m$

$x_{4} = -1.5 + 3.0\times 4-14\times 4^2+5\times 4^3= 106.5 m$

(b)Object 's displacement between between t=2s and t= 4s,, S= x₄ - x₂

(c) Average velocity between t=2s and t= 4s ,, v_av = Total displacement / total time taken

$v_{av}= \frac{118}{2}= 59 m/s$

(d) Instantaneous velocity is taken as v = dx/ dt

so we differntiate x

$v =\frac{\mathrm{d} x}{\mathrm{d} t}=3.0 - 28 t+ 15t^2$

$v_{1} =3.0 - 28 \times 1+ 15\times 1^2= -10 m/s$

$v_{2} =3.0 - 28 \times2+ 15\times 2^2= 7 m/s$

$v_{3} =3.0 - 28 \times3+ 15\times 3^2= 54 m/s$

$v_{4} =3.0 - 28 \times4+ 15\times 4^2= 131 m/s$

(e) Average accelartion between t=2s and t= 4s

a_av = change in velocity / time duration

$a_{av}= \frac{v_{4}-v_{2}}{4-2}=\frac{131 - 7}{2}= 62 m/s^2$

Answer 3

(a) Positions of the object at t = 1, 2, 3,4 seconds are taken as x_1, x₂ , x₃ ,x₄

$x_{1} = -1.5 + 3.0\times 1-14\times 1^2+5\times 1^3= - 7.5 m$

$x_{2} = -1.5 + 3.0\times 2-14\times 2^2+5\times 2^3= -11.5 m$

$x_{3} = -1.5 + 3.0\times 3-14\times 3^2+5\times 3^3= 16.5 m$

$x_{4} = -1.5 + 3.0\times 4-14\times 4^2+5\times 4^3= 106.5 m$

(b)Object 's displacement between between t=2s and t= 4s,, S= x₄ - x₂

(c) Average velocity between t=2s and t= 4s ,, v_av = Total displacement / total time taken

$v_{av}= \frac{118}{2}= 59 m/s$

(d) Instantaneous velocity is taken as v = dx/ dt

so we differntiate x

$v =\frac{\mathrm{d} x}{\mathrm{d} t}=3.0 - 28 t+ 15t^2$

$v_{1} =3.0 - 28 \times 1+ 15\times 1^2= -10 m/s$

$v_{2} =3.0 - 28 \times2+ 15\times 2^2= 7 m/s$

$v_{3} =3.0 - 28 \times3+ 15\times 3^2= 54 m/s$

$v_{4} =3.0 - 28 \times4+ 15\times 4^2= 131 m/s$

(e) Average accelartion between t=2s and t= 4s

a_av = change in velocity / time duration

$a_{av}= \frac{v_{4}-v_{2}}{4-2}=\frac{131 - 7}{2}= 62 m/s^2$