Time & Distance Questions and Answers

Answer: D

The average velocity is 0.

Average velocity is total displacement divided by total time. Ravi runs 90 m to B and returns to A, so he ends where he began and his displacement is zero:

\[ \begin{aligned} v_{avg} &= \frac{\text{displacement}}{\text{time}} \\ &= \frac{0}{60} = 0 \end{aligned} \]

The 3 m/s option is his average speed, 180 m in 60 s, which uses distance instead of displacement. 2 m/s and 1 m/s match neither quantity.

Answer: C

The average speed was 74 km/hr.

The train left at 17:00 on 30 January and arrived at 17:00 on 31 January, so the journey took exactly 24 hours.

\[ \begin{aligned} \text{average speed} &= \frac{\text{distance}}{\text{time}} \\ &= \frac{1776}{24} = 74\ \text{km/hr} \end{aligned} \]

The trap is to treat the equal clock times as zero elapsed time, or to miscount the hours across midnight. The other options correspond to journey times of about 23, 25 and 22 hours, none of which matches the timetable.

Answer: C

The average speed was 53 km/h.

From 19:45 on 29 November to 19:45 on 30 November is 24 hours; to 11:45 on 1 December is another 16 hours, so the journey took 40 hours:

\[ \begin{aligned} \text{speed} &= \frac{2120\ \text{km}}{40\ \text{h}} \\ &= 53\ \text{km/h} \end{aligned} \]

The option 132.5 km/h uses only 16 hours, forgetting the full day in between, 60 km/h does not follow from the times, and 45 km/h would need about 47 hours.

Answer: B

The total distance is 22 km.

Distance = speed × time, with the time converted to hours:

\[ \begin{aligned} d_{1} &= 50 \times \frac{12}{60} = 10\ \text{km} \\ d_{2} &= 40 \times \frac{18}{60} = 12\ \text{km} \\ d &= 10 + 12 = 22\ \text{km} \end{aligned} \]

Averaging the two speeds (45 km/h) over the 30 minutes gives 22.5 km, which is wrong because the bus spends different times at each speed. Swapping the two times gives 15 + 8 = 23 km, and 20, 24 and 28 km do not follow from distance = speed × time with the times given.

Answer: B

The average speed is 15 m/s.

Convert the distance to metres and the time to seconds, then divide.

\[ \begin{aligned} 756 \text{ km} &= 756000 \text{ m} \\ 14 \text{ h} &= 14 \times 3600 = 50400 \text{ s} \\ \text{speed} &= \frac{756000}{50400} = 15 \text{ m/s} \end{aligned} \]

The speed in km/h is \(756 \div 14 = 54\), and 54 m/s is the trap of forgetting to convert it: \(54 \times \frac{5}{18} = 15\). 900 is the speed in metres per minute, \(54000 \div 60\), and 30 m/s doubles the true value.

Answer: A

The average speed is 20 m/s.

\[ \begin{aligned} \text{speed} &= \frac{792\ \text{km}}{11\ \text{h}} = 72\ \text{km/h} \\ 72 \times \frac{5}{18} &= 20\ \text{m/s} \end{aligned} \]

To convert km/h to m/s, multiply by \(\frac{1000}{3600} = \frac{5}{18}\).

72 m/s is the speed left in km/h with the wrong unit attached, which is the main trap. 36 m/s and 15 m/s come from conversion errors.

Answer: B

The average speed was 63 km/h.

The train left at 17:00 on 30 April and arrived at 20:00 on 1 May:

\[ \begin{aligned} \text{time} &= 7 + 20 = 27\ \text{h} \\ \text{speed} &= \frac{1701}{27} = 63\ \text{km/h} \end{aligned} \]

The 7 hours run from 17:00 to midnight and the 20 hours from midnight to 20:00. 62.7 km/h and 63.3 km/h are what you get with a time slightly too long or too short, and 60 km/h does not come from 27 hours.

Answer: B

Her average speed is 9.6 km/h.

Average speed is total distance divided by total time, not the average of the two speeds:

\[ \begin{aligned} t &= \frac{5}{12} + \frac{5}{8} = \frac{10 + 15}{24} = \frac{25}{24}\ \text{h} \\ v &= \frac{10}{25/24} = \frac{240}{25} = 9.6\ \text{km/h} \end{aligned} \]

For equal distances this is the harmonic mean, \(\frac{2 \times 12 \times 8}{12 + 8} = 9.6\). 10 km/h is the simple average of 12 and 8, the common mistake, because she spends longer at the slower speed. 20 km/h adds the speeds, and 5 m/s (18 km/h) is far too fast.

Answer: C

XZ is 53 1/3 km. A and B start together from X. B reaches Y after \[\frac{60}{20} = 3\] hours, by which time A has covered 16 x 3 = 48 km, so A is 12 km from Y. B now comes back towards A and they approach each other at 16 + 20 = 36 km/h. They meet after \[\frac{12}{36} = \frac{1}{3}\] hour, in which A goes a further 16/3 = 5 1/3 km. Hence \[XZ = 48 + 5\tfrac{1}{3} = 53\tfrac{1}{3}\] km. The other options, 52 1/2, 52 1/4 and 55 2/3, do not satisfy these distances.

Answer: D

The journey is 8 km. Let the distance be \(d\) km. The time at 20 km/h exceeds the time at 80 km/h by 18 minutes, which is \(0.3\) h: \[ \frac{d}{20} - \frac{d}{80} = 0.3 \] \[ \frac{4d - d}{80} = 0.3 \] \[ d = \frac{0.3\times 80}{3} = 8 \] Check: 8 km at 80 km/h takes 6 minutes and at 20 km/h takes 24 minutes, a difference of 18 minutes. The values 6, 10 and 18 km do not give that difference.

A shortcut: for one distance at two speeds the time gap is \(d\left(\tfrac{1}{20} - \tfrac{1}{80}\right) = \tfrac{3d}{80}\), and setting it equal to 0.3 hour gives d = 8 km.