Quantitative Aptitude

Time, Speed and Distance

2,165 Questions
All Core formulas and unit conversion 441 Boats and streams effective speed 396 Relative speed, meetings, and overtaking 375 Average speed 362 Early and late arrival 321 Proportional relationships 202 Stoppage time per hour 35 Train and stationary or moving object crossing 17 Races and dead heats 16

Time, Speed and Distance Questions

Multiple choice

Two motorcycles $M_{1}$ and $M_{2}$ are heading towards each other with a speed of $30 kmh^{-1}$ each. A bird flies off $M_{1}$ at $60 kmh^{-1}$ when distance between the motorcycles is $60 km$. It heads towards $M_{2}$ and then back to $M_{1}$ and so on. The total distance the bird moves till the motorcycles meet is
  1. $60 \:km$
  2. $40 \:km$
  3. $50 \:km$
  4. $30 \:km$

  5. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The motorcycles are 60 km apart and each moves at 30 km/h, so they approach each other at a relative speed of 60 km/h. They will meet in 60 km / 60 km/h = 1 hour. Since the bird flies at a constant speed of 60 km/h for the entire hour, it covers 60 km/h * 1 h = 60 km.

Multiple choice

Time interval (in sec) Speed of particle (in m/s) 0-2 2 2-5 3 5-10 4 10-15 2 A table is given below of a particle moving along x-axis. In the table speed of particle at different time intervals is shown. Find total distance travelled by the particle and its average speed.

  1. 43 m, 2.87 m/s

  2. 40 m, 2.66 m/s

  3. 43 m, 2.66 m/s

  4. 40 m, 2.87 m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance = speed * time. Distances: (2*2) + (3*3) + (4*5) + (2*5) = 4 + 9 + 20 + 10 = 43m. Total time = 2+3+5+5 = 15s. Average speed = 43/15 = 2.866... m/s.

Multiple choice

A car is moving at a constant speed of 40 km/h along a straight road which heads towards a large vertical wall and makes sharp $90^0$ turn by side of the wall. A fly flying at a constant speed of 100 km/h , starts from the wall towards the car at an instant when the car is 20 km away, flies until it reaches the glasspane of the car and returns to the wall at the same speed. It continues to fly between the car and the wall toll the car makes the $90^0$ turn. What is the total distance the fly has travelled during this period?

  1. 20 km

  2. 30 km

  3. 50 km

  4. 70 km
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The car travels 20 km at 40 km/h, taking 0.5 hours. During this time, the fly flies at 100 km/h. Total distance = speed * time = 100 * 0.5 = 50 km.

Multiple choice

Two trains start a distance of $2000m$ apart. Train one is moving with a constant speed of $30 m/s$ directly towards train $2$ which starts from rest and accelerates with a constant acceleration of $5m/s^2$ directly towards train 1. When do the trains meet?
  1. $22.9 s$
  2. $34.9 s$
  3. $30 s$
  4. $40 s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Relative acceleration = 5 m/s^2. Initial relative velocity = 30 m/s. Distance = 2000 m. Using s = ut + 0.5at^2: 2000 = 30t + 0.5(5)t^2. 2.5t^2 + 30t - 2000 = 0. t^2 + 12t - 800 = 0. Using quadratic formula: t = (-12 + sqrt(144 + 3200)) / 2 = (-12 + 57.8) / 2 = 22.9 s.

Multiple choice

The driver of a train moving with a constant speed $v_1$ along a straight track sights another train at a distance d ahead of him on the same track moving in the same direction with a constant speed $v_2$. He at once applies the brakes and gives his train a constant retardation f. There will be a collision of the trains if:
  1. $\displaystyle v_1 > v_2$ and $\displaystyle \frac{(v_1 - v_2)^2}{2f} < d$
  2. $\displaystyle v_1 < v_2$ and $\displaystyle \frac{(v_1 + v_2)^2}{2f} > d$
  3. $\displaystyle v_1 > v_2$ and $\displaystyle \frac{(v_1 - v_2)^2}{2f} > d$
  4. $\displaystyle v_1 > v_2$ and $\displaystyle \frac{(v_1^2 - v_2^2)}{2f} > d$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The relative speed is (v1 - v2). The distance covered by the first train to stop is v1^2 / (2f). The distance covered by the second train in that time is v2 * (v1/f) - 0.5 * f * (v1/f)^2? No, simpler: relative velocity v_rel = v1 - v2. Relative acceleration a_rel = -f. Collision occurs if the relative stopping distance (v1-v2)^2 / (2f) is greater than the initial distance d.

Multiple choice

A man running with a uniform speed 'u' on a straight road observes a stationary bus at a distance 'd' ahead of him. At that instance , the bus starts with an acceleration 'a' .The condition that he would be able to catch the bus is:
  1. $\displaystyle d \, \leq \, \frac{u^2}{a}$
  2. $\displaystyle d \, \leq \, \frac{u^2}{2a}$
  3. $\displaystyle d \, \leq \, \frac{u^2}{3a}$
  4. $\displaystyle d \, \leq \, \frac{u^2}{4a}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The man catches the bus if his distance covered in time t is greater than or equal to the bus's distance plus the initial gap d. ut >= 0.5at^2 + d. Rearranging: 0.5at^2 - ut + d <= 0. For a real solution for t, the discriminant must be >= 0. (-u)^2 - 4(0.5a)(d) >= 0, so u^2 - 2ad >= 0, which means d <= u^2 / 2a.

Multiple choice

A bird is sitting on train A moving towards East with a velocity 300 km/hr. Another train B of same speed is moving in West direction on the same track. When the trains are 6 km apart, the bird starts flying with a velocity 30 km/hr with respect to ground towards B. After touching B, it returns back to A and continue repeating this process until the trains collide. In this process, the total distance traveled by the bird is:

  1. 100 m

  2. 200 m

  3. 300 m

  4. 6 km
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The trains are 6 km apart and moving at 300 km/hr each. Relative speed = 600 km/hr. Time to collide = 6 km / 600 km/hr = 0.01 hours = 36 seconds. The bird flies at 30 km/hr for 36 seconds. Distance = 30 * (36/3600) = 0.3 km = 300 m.

Multiple choice

A boat takes two hours to travel 8 km down and 8 km up the river when the water is still. How much time will the boat take to make the same trip when the river starts flowing at 4 kmph?

  1. 2 hour

  2. 2 hour 40 minute

  3. 3 hour

  4. 3 hour 40 minute
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In still water, the boat takes 2 hours for a total of 16 km, meaning its speed is 8 km/h. When the river flows at 4 km/h, the downstream speed is 12 km/h and the upstream speed is 4 km/h. The total time for the trip is 8/12 + 8/4 = 2/3 + 2 = 2 hours and 40 minutes.

Multiple choice

A boat takes $4$ hours to move $10\ km$ and back to starting point in still water. If the river flow velocity is $2\ km/hr$, then time taken by the boat to move $10\ km$ upstream and back to starting point is approximately
  1. $4\ hours$
  2. $5\ hours$
  3. $6\ hours$
  4. $7\ hours$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In still water, the boat travels 10 km and back in 4 hours, so speed in still water (v) is 10 / 2 = 5 km/h. With stream speed (u) = 2 km/h, upstream speed is 5-2 = 3 km/h and downstream is 5+2 = 7 km/h. Total time = 10/3 + 10/7 = 3.33 + 1.43 = 4.76 hours, which is approximately 5 hours.