Question

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?

Answer 1

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 2

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

$v = u+at$

$t=\dfrac{v-u}{a}$

Where, v = final velocity

u = initial velocity

t = time

a = acceleration

Put the value into the formula

$t =\dfrac{0-7}{-2.5}$

$t=2.8\ sec$

We need to calculate the distance

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

$s=7\times2.8-\dfrac{1}{2}\times2.5\times(2.8)^2$

$s=9.8\ m$

Hence, The distance is 9.8 m.