Speed, Distance and Time Worked ExamplesIGCSE Maths 0580

(a) The train arrives at 23 52 and leaves at 00 04. Count up to midnight first, then past it.

Add the two parts.

(b) From 21 45 to midnight is $2\text{ h }15\text{ min}$; from midnight to 04 15 is $4\text{ h }15\text{ min}$.

(c) The distance between the two stations is the difference of their distances from Aldon.

The train leaves Brenmoor at 00 04 and arrives at Carlow at 02 34, so it takes $2\text{ h }30\text{ min} = 2.5\text{ h}$. Use hours because the answer is in km/h.

Double top and bottom to divide easily without a calculator.

(a) First check the division is right. $70 \times 2 = 140$ and $70 \times 0.35 = 24.5$, and $140 + 24.5 = 164.5$, so $2.35$ hours is correct.

The mistake is in the conversion. $0.35$ is a decimal fraction of an hour, not $35$ minutes — there are $60$ minutes in an hour, not $100$.

(b) Convert the decimal part to minutes by multiplying by $60$.

Add to the departure time: 13 50 plus $2$ hours is 15 50, and $21$ more minutes takes it past the hour to 16 11.

(a) The lengths are in metres and the time is wanted in seconds, so convert the speed to m/s. $1\text{ km/h}$ is $1000\text{ m}$ in $3600\text{ s}$.

From the front entering to the back leaving, the front of the train travels the length of the tunnel plus the length of the train.

Divide distance by speed.

(b) The whole train is inside from the moment its back enters until its front reaches the far end. In that time the front moves the tunnel length minus the train length.

At $20\text{ m/s}$:

(c) Work backwards. The second train's front travels $1020 + 240 = 1260\text{ m}$ in $50\text{ s}$.

Convert back to km/h by multiplying by $3.6$ (that is, $\times 3600 \div 1000$).

(a) Put both times in the same time zone. When it is 23 40 in Singapore, London is $7$ hours behind, so the plane leaves at 16 40 London time.

Now count from 16 40 to 06 25 the next day, both London time: $7\text{ h }20\text{ min}$ to midnight, then $6\text{ h }25\text{ min}$.

(b) Enter the time as a decimal: $45$ minutes is $\frac{45}{60} = 0.75$ of an hour.

Divide distance by time and round to $3$ significant figures.

(c) The flight time is distance divided by speed. The calculator displays $12.8$.

Interpret the display: $0.8$ of an hour is $0.8 \times 60 = 48$ minutes.

Add to the London departure time: 21 15 plus $12\text{ h }48\text{ min}$ is 10 03 the next day, London time. Singapore is $7$ hours ahead, so add $7$ hours.

(a) The two semicircles together make one full circle of radius $35\text{ m}$.

Add the two straights. Show more figures than the value you are asked to show.

(b) Use the unrounded lap length. Twelve laps in kilometres:

Convert the time to hours: $12$ seconds is $0.2$ minutes.

Divide and round to $3$ significant figures.

(c)(i) Pace is time per unit distance — the reverse of speed. $4\text{ min }30\text{ s} = 4.5$ minutes for $1\text{ km}$, so count kilometres in $60$ minutes.

(ii) Time $=$ distance $\times$ pace.

Convert the decimal part of a minute to seconds.

(a) Ama goes from $0\text{ km}$ at 09 00 to $6\text{ km}$ at 10 30, which is $1.5$ hours.

(b) A horizontal line means the distance from $A$ is not changing. Ama's line is flat from 10 30 to 11 00.

(c) Ben leaves $B$ ($12\text{ km}$) at 10 15 and reaches $A$ ($0\text{ km}$) at 11 15 — $12\text{ km}$ in $1$ hour. The line slopes down only because he is travelling towards $A$; speed is positive.

(d) They pass where the two lines cross. Ben's line crosses Ama's flat section at $6\text{ km}$, halfway between 10 30 and 11 00. Check: Ben has covered $12 - 6 = 6\text{ km}$ at $12\text{ km/h}$, which takes $30$ minutes after 10 15.

(e) Read both graphs at 11 00. Ama is still at $6\text{ km}$. Ben has been riding for $45$ minutes, covering $12 \times 0.75 = 9\text{ km}$, so he is $3\text{ km}$ from $A$ — as the graph shows.

(a) Time for each driving leg, in minutes. $A$ to $B$:

$B$ to $C$:

Add the two legs and the stop to the start time.

(b) From 13 20 to 14 00 leaves $40$ minutes $= \frac{2}{3}$ hour for the $72\text{ km}$ from $C$ to $D$.

(c) $108 > 100$, so she cannot arrive on time legally. At $100\text{ km/h}$ the leg takes:

Interpret $0.72$ hours: $0.72 \times 60 = 43.2$ minutes, and $0.2 \times 60 = 12$ seconds.

She had $40$ minutes available, so she is late by the difference (she arrives at 14 03 12).

(a) Standing still, Mei moves at the walkway's speed.

(b) Walking with the walkway, her speed over the ground is her walking speed plus the walkway speed.

Solve for $w$.

(c) Against the walkway the speeds subtract. The child makes progress only because $1.75 > 1.25$.

Time for $60\text{ m}$:

(d) Time for each half: walking at $3\text{ m/s}$ over the ground, then standing at $1.25\text{ m/s}$.

Average speed is total distance divided by total time.

(a) Time $=$ distance $\div$ speed. His downhill speed is $(v + 6)\text{ km/h}$, so write each time as a fraction.

The two times add up to $1$ hour.

Multiply every term by $v(v + 6)$ to clear the fractions.

Expand, then collect every term on one side.

(b) Factorise: two numbers that multiply to $-36$ and add to $-9$ are $-12$ and $3$.

A speed cannot be negative, so $v = 12$. The question asks for the downhill speed, which is $6\text{ km/h}$ faster.

(c) Uphill he rides $6\text{ km}$ at $12\text{ km/h}$. Check: downhill takes $\frac{9}{18} = \frac{1}{2}$ hour, and the two halves make $1$ hour.

(a) Write each time as distance $\div$ speed, and change $3$ hours $30$ minutes to $3.5$ hours.

Multiply every term by $2(v - 3)(v + 3)$, which clears both fractions and the decimal.

Expand each bracket. The constant terms on the left cancel.

Move everything to the right-hand side. Every line is shown because this is a show that.

(b) The quadratic does not factorise, so use the formula with $a = 7$, $b = -72$ and $c = -63$.

Evaluate both roots and round to $2$ decimal places.

A speed cannot be negative (and the boat must be faster than the $3\text{ km/h}$ current to get upstream at all), so the speed is $11.10\text{ km/h}$.

(c) Upstream speed $= v - 3$. Use the unrounded value of $v$.

Convert the decimal part of the hour to minutes, then round.