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A baseball player is stealing second base when he goes into a slide, which gives him an acceleration of −2.5 m/s2. If he had an initial speed of 7 m/s, how far does he slide before coming to a complete stop?

Respuesta :

sashakat
first, find Δt
Δt= velocity÷acceleration
⇔(0-7)÷(-2.5)
⇔2.8 seconds.

replace it in ΔX=1/2aΔt^2+viΔt
⇔ ΔX= 1/2(-2.5)(2.8)^2+7(2.8)
⇔9.8m

Answer:

The distance is 9.8 m.

Explanation:

Given that,

Acceleration = -2.5 m/s²

Initial speed = 7 m/s

We need to calculate the time

Using equation of motion

[tex]v = u+at[/tex]

[tex]t=\dfrac{v-u}{a}[/tex]

Where, v = final velocity

u = initial velocity

t = time

a = acceleration

Put the value into the formula

[tex]t =\dfrac{0-7}{-2.5}[/tex]

[tex]t=2.8\ sec[/tex]

We need to calculate the distance

Using equation of motion

[tex]s=ut+\dfrac{1}{2}at^2[/tex]

[tex]s=7\times2.8-\dfrac{1}{2}\times2.5\times(2.8)^2[/tex]

[tex]s=9.8\ m[/tex]

Hence, The distance is 9.8 m.

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