Question

A skateboard chsnges velocity from 15 meters per second to 5 meters per second, over the course of 5 seconds. What is the skateboards acceleration?

A. -2 m/s^2
B. +2 m/s^2
C. +0.5 m/s^2
D. -0.5 m/s^2​

Answer 1

-2 m/s²

Step-by-step explanation:

Here it's given that a skateboard changes its velocity from 15m/s to 5m/s in an interval of 5s . We are interested in finding out the acceleration of the skateboard. Firstly we know that,

• The rate of change of velocity is called acceleration. Mathematically,

`\qquad\rm\twoheadrightarrow accl^n =(v - u )/(t)`

Here ,

• initial velocity = 15m/s
• final velocity = 5m/s
• time = 5s

On substituting the respective values, we have;

`\qquad\rm\twoheadrightarrow accl^n =(5m/s-15m/s)/(5s)`

Simplify numerator,

`\qquad\rm\twoheadrightarrow accl^n =(-10m/s)/(5s)`

Divide,

`\qquad\rm\twoheadrightarrow \underline{\boxed{\blue{\rm acceleration= -2\ m/s^2}}}`

Therefore, Option A is correct choice.

Answer 2

Option A) is correct.

We are given –

• Initial velocity of skateboard is, u= 15m/s
• Final velocity of skateboard is,v = 5 m/s
• Time taken, t = 5 sec
• Acceleration, a =?

Formula To be used:-

`\qquad`
`\star\: \: \pink{\boxed{\frak{ a = ((v-u))/(t)}}}`

Where–

• A = Acceleration
• v = Final velocity
• u = Initial velocity
• t = time

`\qquad\qquad\quad\underline{\sf{Substituting \ Values \ :}}`

⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━

`\qquad`
`\bf \longrightarrow a = (v-u)/(t)`

`\qquad`
`\sf \longrightarrow a = (5-15)/(t)`

`\qquad`
`\sf \longrightarrow a = (-10)/(5)`

`\qquad`
`\sf \longrightarrow a = \cancel{(-10)/(5)}`

`\qquad`
`\pink{ \bf \longrightarrow a = -2 \:m/s^2}`

• Henceforth, acceleration is -2 m/s².