Question

Abigail was skateboarding home when the wheel axle of her skateboard broke. She had already traveled two thirds of the way home and had to walk the rest of the way. Walking the rest of the way home took her twice as long as it took her to ride her skateboard. How many times faster is Abigail on her skateboard than she is walking?

Answer 1

The number of times faster Abigail is on her skateboard than she is walking is 4 times

Explanation:

The information given are;

Abigail was skateboarding home (let the distance home = D)

The distance Abigail had traveled when her skateboard broke = 2/3 of D

The distance of the rest (remaining) of the way home = 1 - 2/3 of D = 1/3 of D

The time Abigail took in walking the rest of the way home = 2 × t, Time it will take on the skateboard

Given that t is the time it takes Abigail to arrive home on skateboard alone, we have;

`Speed = (Distnce)/(Time)`

Speed on skate board, S
`_t` = Distance/Time = D/t

Therefore, the time it took Abigail to travel two thirds (2/3) of D is given as follows;

`Time= (Distnce \ covered)/(Speed ) = ((2)/(3) D)/(S_t) = ((2)/(3) D)/((D)/(t) ) = (2)/(3) D * (t)/(D) = (2)/(3) t`

The total time it took Abigail to get to her home = Time of walking + Time on skateboard

The question also states that he total time it took Abigail to get to her home by walking and skateboarding because her skateboard broke = 2×t

Therefore, the total time it took Abigail to get to her home = 2×t = Time of walking + Time on skateboard = Time of walking +
`(2)/(3) t`

Which gives;

2×t = Time of walking +
`(2)/(3) t`

Time of walking =
`2 * t -(2)/(3) t` =
`(4)/(3) t`

∴ The time Abigail walked 1/3·D =
`(4)/(3) t`

Abigail walking speed is then

`Walking \ speed= (Distnce \ covered \ walking)/(Time \walking) = ((1)/(3) D)/((4)/(3) t) = ((D)/(3) )/((4 \cdot t)/(3) ) = (D)/(3) } * {(3)/(4 \cdot t) } = (D)/(4 \cdot t) }`

To compare how many times faster is Abigail on her skateboard than she is walking, we divide the expression (formula) for the speed of skateboarding with the (formula) for the speed of walking as follows;

`The \ number \ of \ times \ faster = ((D)/(t) )/((D)/(4 \cdot t) )} = (D)/(t) * (4 \cdot t)/(D) } = 4`

The number of times faster is 4 that is Abigail is four times faster on her skateboard than she is walking.