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A blue shark accelerates at a rate of 16m/s for a time of 0.8s. During this time it travels a
distance of 11.52m. Calculate its initial speed.

Respuesta :

jacob193

Answer:

The initial velocity of this shark is [tex]8.0\; \rm m \cdot s^{-1}[/tex]. (Assuming that the unit of the acceleration in this question is [tex]\rm m\cdot s^{-2}[/tex].)

Explanation:

Let [tex]a[/tex] denote the acceleration of this shark.

Let [tex]v_0[/tex] denote the initial velocity of this shark.

Assume that the acceleration [tex]a[/tex] of this shark is constant (as it is in this question.) Over a period of time [tex]t[/tex], the shark would have travelled a distance of:

[tex]\displaystyle x = \frac{1}{2}\, a\, t^2 + v_0\, t[/tex].

This question states that:

  • [tex]x = 11.52\; \rm m[/tex],
  • [tex]t = 0.8\; \rm s[/tex]. (That is: this shark travelled a distance of [tex]11.52\; \rm m[/tex] in [tex]0.8\; \rm s[/tex].)
  • [tex]a = 16\; \rm m \cdot s^{-2}[/tex] (the acceleration of this shark is indeed a constant.)

This question is asking for [tex]v_0[/tex], the initial velocity of this shark at the beginning of this [tex]0.8[/tex]-second period. Substitute the three known values into the equation:

[tex]\displaystyle 11.52 = \frac{1}{2}\times 16\times (0.8)^2 + 0.8\, v_0[/tex].

Solve for [tex]v_0[/tex]:

[tex]v_0 = \displaystyle \frac{11.52 - (1/2) \times 16 \times (0.8)^2}{0.8} = 8.0\; \rm m \cdot s^{-1}[/tex].

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