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

The distance is 9.8 m.

Step-by-step 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=(v-u)/(a)`

Where, v = final velocity

u = initial velocity

t = time

a = acceleration

Put the value into the formula

`t =(0-7)/(-2.5)`

`t=2.8\ sec`

We need to calculate the distance

Using equation of motion

`s=ut+(1)/(2)at^2`

`s=7*2.8-(1)/(2)*2.5*(2.8)^2`

`s=9.8\ m`

Hence, The distance is 9.8 m.