Time Speed and Distance Aptitude Solved Questions:

Answer: (b) \(60 \ km/hr\)

Solution: Given, Initial speed of the train \((v_i) = 45 \ km/hr\)

final speed of the train \((v_f) = 75 \ km/hr\)

time taken by the train to achieve the final speed \((T) = 0.5 \ hours\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ = \frac{75 - 45}{0.5} $$ $$ = \frac{30}{0.5} = 60 \ km/hr $$

Answer: (a) \(20 \ km/hr\)

Solution: Given, initial speed of the car \((v_i) = 25 \ km/hr\)

final speed of the car \((v_f) = 65 \ km/hr\)

time taken by the car to achieve final speed \((T) = 2 \ hours\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ = \frac{65 - 25}{2} $$ $$ = \frac{40}{2} = 20 \ km/hr $$

Answer: (b) \(2 \ hours\)

Solution: Given, Initial speed of the man \((v_i) = 60 \ km/hr\)

final speed of the man \((v_f) = 120 \ km/hr\)

Acceleration \((a) = 30 \ km/hr\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 30 = \frac{120 - 60}{T} $$ $$ T = \frac{60}{30} = 2 \ hours $$

Answer: (d) \(187.5 \ km/hr\)

Solution: Given, total distance covered by the train = \(800 \ km + 700 \ km\) = \(1500 \ km\)

total time taken by the train \(= 3 + 5 = 8 \ hours\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{1500}{8} = 187.5 \ km/hr $$

Answer: (a) \(130 \ m/sec\)

Solution: Given, initial speed of the man \((v_i) = 10 \ m/sec\)

time taken by the man to achieve the final speed \((T) = 60 \ seconds\)

acceleration rate of the bicycle \((a) = 2 \ m/sec\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 2 = \frac{v_f - 10}{60} $$ $$ 120 = v_f - 10 $$ $$ v_f = 130 \ m/sec $$

Answer: (c) \(102.996 \ km/hr\)

Solution: Given, final speed of the bus \((v_f) = 175 \ km/hr\) = \(175 \times \frac{5}{18}\)

time taken by the bus to achieve the final speed \((T) = 2 \ seconds\)

rate of accleration of the bus \((a) = 100 \ m/sec\), then $$ Acceleration \ (a) = \frac{v_f - v_i}{T} $$ $$ 10 = \frac{175 \times \frac{5}{18} - v_i}{2} $$ $$ 20 \times 18 = 175 \times 5 - 18 \ v_i $$ $$ 18 \ v_i = 515 $$ $$ v_i = 28.61 \ m/sec $$ $$ v_i = 28.61 \times \frac{18}{5} \ km/hr $$ $$ v_i = 102.996 \ km/hr $$

Answer: (d) \(12.5 \ km\)

Solution: Given, average speed of the man = \(500 \ m/sec\)

time taken by the man to cover the distance = \(25 \ seconds\), then $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ 500 = \frac{distance \ covered}{25} $$ $$ distance \ covered = 12,500 \ meter $$ $$ = 12,500 \times \frac{1}{1000} = 12.5 \ km $$

Answer: (b) \(260 \ km/hr\)

Solution: Given, speed of the first car \((s_1) = 150 \ km/hr\)

speed of the second car \((s_2) = 110 \ km/hr\), then $$ Relative \ speed = s_1 + s_2 $$ $$ = 150 + 110 = 260 \ km/hr $$

Answer: (a) \(85.7 \ km/hr\)

Solution: $$ Average \ speed = \frac{distance \ covered}{time \ taken} $$ $$ = \frac{d_1 + d_2}{\frac{d_1}{s_1} + \frac{d_2}{s_2}} $$ $$ = \frac{200 + 200}{\frac{200}{75} + \frac{200}{100}} $$ $$ = \frac{400 \times 300}{800 + 600} $$ $$ = \frac{120000}{1400} = 85.7 \ km/hr $$

Answer: (b) \(36 \ km/hr\)

Solution: speed of the girl = \(10 \ m/sec\), then $$ = 10 \times \frac{18}{5} \ km/hr $$ $$ = 36 \ km/hr $$