Question

1) a) A particle’s velocity is given (in units of meters per second) by the function
v = 1 + (t^3)/10 – (t^2)/20
Find the distance travelled from t = 0 and t = 2.

Input your final answer (the value of each side)

b)Explanation:
Assume the reader understands derivatives, and knows the definition of instantaneous velocity (dx/dt), but is new to integrals and is struggling to understand them. Use students’ prior knowledge to provide an explanation that includes the concept and physical meaning of the integral of velocity with respect to time.

Answer 1

distance travelled from t = 0 and t =  2   = 2.267  if v = 1  + t³/10  - t²/20

Step-by-step explanation:

V = dx/dt

dx/dt = 1  + t³/10  - t²/20

∫ dx  = ∫ (1  + t³/10  - t²/20) dt

=> x =  t  + t⁴/40  - t³/60  + c

c is constant

Distance  traveled from t = 0  & t = 2

=> ( 2 + 2⁴/40  - 2³/60 + c )  - (0  + 0 - 0 + c)

= 2 + 16/40 - 8/60

= 2 + 0.4 - 0.133

= 2.267

distance travelled from t = 0 and t =  2   = 2.267

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Answer 2

Answer:

Step-by-step explanation:

$velocity, v = \frac{\mathrm{d} x}{\mathrm{d} t}\\v = 1+\frac{t^{3}}{10}-\frac{t^{2}}{20}$

$dx = 1.dt + \frac{t^{3}}{10} dt -\frac{t^{2}}{20} dt$

Integrating on both sides

$\int\,dx= \int\ {1} \, dt + \int \frac{t^{3}}{10}dt -\int \frac{t^{2}}{20}dt$

Implies

$x = t +\frac{t^{4}}{40} - \frac{t^{3}}{60}+ c$

displacement at t = 0 ,

$x_0 = 0 + \frac{0^{4}}{40} - \frac{0^{3}}{60} + c = c$

displacement  at t= 2,

$x_2 = 2 + \frac{2^{4}}{40} - \frac{2^{3}}{60} + c$

$x_2 = 2 + \frac{16}{40} - \frac{8}{60} + c$

$x_2 = \frac{34}{15} + c$

displacement from t = 0 to t = 2 is $x_2 - x_0$ ,

$x_2 - x_0 = \frac{34}{15} + c - c$

            $= 2.267$ m

Therefore , the particle travelled 2.267 m  from t = 0 to t  = 2