Model 6 Change in speed Vs Change with time travel Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Let the distance be x km and initial speed be y kmph.

According to question,

$x/y - x/{y + 3} = 40/60$ ...(i)

and, $x/{y - 2} - x/y = 40/60$ ...(ii)

From equations (i) and (ii),

$x/y - x/{y + 3} = x/{y - 2} - x/y$

$1/y - 1/{y + 3} = 1/{y - 2} - 1/y$

${y + 3 -y}/{y(y + 3)} = {y - y + 2}/{y(y - 2)}$

3 (y - 2) = 2 (y + 3)

3y - 6 = 2y + 6

y = 12

From equation (i),

$x/12 - x/15 =40/60$

= ${5x - 4x}/60 = 2/3$

$x = 2/3 × 60$ = 40

Distance = 40 km.

Answer: (a)

Using Rule 1,

Speed of train = $\text"Distance"/\text"Time"$

= $10/{12/60}$ kmph

= ${10 × 60}/12 = 50$ kmph

New speed = 45 kmph

∴ Required time = $10/45$ hour

= $2/9 × 60$ minutes

= $40/3$ minutes or 13 minutes 20 seconds

Answer: (a)

Let the required distance be x km.

$x/{5/2} - x/3 = 16/60$

${2x}/5 - x/3 = 4/15$

${6x - 5x}/15 = 4/15 ⇒ x = 4$ km.

Using Rule 10,

Here, $S_1 = 2{1}/2 , t_1 = 6, S_2 = 3, t_2$ = 10

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${5/2 × 3(6 + 10)}/{3 - 5/2}$

= $15 × 16/60$ km = 4 km

Answer: (b)

Let the distance be x km.

$x/10 - x/12 = 12/60$

${6x - 5x}/60 = 1/5$

$x = 1/5 × 60$ = 12 km.

Using Rule 10,

Here, $S_1 = 10, t_1 = 6, S_2 = 12, t_2$ = 6

Distance = ${(S_1 × S_2)(t_1 + t_2)}/{S_2 - S_1}$

= ${(10 × 12)(6 + 6)}/{12 - 10}$

= ${120 × 12}/2$

= 60 × $12/60$ km = 12km

Answer: (b)

If the distance be x km,

then $x/40 - x/50 = 6/60$

$x/4 - x/5 = 1$

x = 20 km.

Required time

= $(20/40)$ hour - 11 minutes

= $(1/2 × 60 - 11)$ minutes

= 19 minutes

Answer: (a)

Let the initial speed of the car be x kmph and the distance be y km.

Then, y = $9/2$ x ...(i)

and, y = 4 (x + 5) ...(ii)

${9x}/2 = 4(x + 5)$

9x = 8x + 40

x = 40 kmph