Question

Help with part b pls thank u

Answer 1

see below

Explanation:

Let t = time (in hours)

a) i) First 105 km of journey

`\mathsf{speed=(distance)/(time)}`

`\implies x=(105)/(t_1)`

`\implies t_1=(105)/(x)`

ii) Rest of journey

distance = 190 - 105 = 85 km

speed =
`x+12` km/h

`\mathsf{speed=(distance)/(time)}`

`\implies x+12=(85)/(t_2)`

`\implies t_2=(85)/(x+12)`

b) 3 hours 15 minutes = 3.25 hours

`\implies` total time =
`t_1+t_2=3.25`

`\implies (105)/(x)+(85)/(x+12) =3.25`

`\implies (105(x+12)+85x)/(x(x+12)) =3.25`

`\implies 105(x+12)+85x =3.25x(x+12)`

`\implies 105x+1260+85x =3.25x^2 +39x`

`\implies 3.25x^2-151x-1260=0`

multiply by 4:

`\implies 13x^2-604x-5040=0`

c)

`\textsf{quadratic formula} \ x=(-b \pm√(b^2-4ac) )/(2a)}`

`\implies x=(604 \pm√((-604)^2-4(13)(-5040)) )/(2(13))}`

`\implies x=(604 \pm√(626896) )/(26)}`

`\implies x = 53.68, x=-7.22`

(to 2 decimal places)

d) time ≥ 0, therefore, x = 53.68 only

Substitute x = 53.68 into the expression for
`t_1` and solve for
`x`:

`\implies t_1=(105)/(53.68)=1.955912... \textsf{hours}=\textsf{1 hr 57 min}`

Answer 2

• See below

Explanation:

• speed = distance / time
• time = distance / speed

a)

i)

Time taken at the first part:

• 105 / x

ii)

Time taken at the second part:

• (190 - 105) / (x + 12) =
• 85/(x + 12)

b)

If time is 3 hours and 15 minutes = 3 1/4 hours = 13/4 hours, the equation is:

• 105/x + 85/(x + 12) = 13/4
• 4(x + 12)*105 + 4(85x) = 13x(x + 12)
• 13x² - 604x - 5040 = 0

c)

Solve the equation:

• 13x² - 604x - 5040 = 0
• x = [604 ± √((-604)² + 4*13*5040) ]/26 =
• [604 ± √626896]/26 =
• [604 ± 791.77]/26
• x = (604 + 791.77)/26 = 53.68 km/h (taking positive root only)

d)

• 105/x = 105/53.68 h = 1.956 h = 1 h 0.956*60 min = 1 h 57 min (rounded)
• 85/(53.68 + 12) = 85/65.68 = 1.294 h = 1 h 0.294*60 min = 1 h 18 min (rounded)