A car, initially at rest, picks up a velocity of 90 km h–1, over a distance of 20 m. Calculate acceleration of car, time in which it picks up above velocity?
| Given Initial velocity of car (u) | = | 0 |
| Final velocity of car (v) | = | 90 km h–1 |
| = | 25 ms–1 | |
| Distance covered by car (S) | = | 20 m. |
| Acceleration (a) | = | ? |
| and Time (t) | = | ? |
| 1) Applying v2 – u2 | = | 2aS. |
| (25)2 – (0)2 | = | 2 × a × 20 |
| ∴ a | = | 625/40 ms–2 |
| = | 15.625 ms–2. |
| 2) Applying v | = | u + at. |
| 25 | = | 0 + 15.625 × t |
| ∴ t | = | 25/15.625 s |
| = | 1.6 s. |
An aeroplane lands with a velocity of 432 km hr–1 and comes to rest, after covering a runway of 1200 m. Calculate retardation and time in which aeroplane comes to rest?
| Given Initial velocity of car (u) | = | 432km hr–1 |
| = | 120 ms–1 | |
| Final velocity of car (v) | = | 0 ms–1 |
| Distance (S) | = | 1200 m. |
| 1) Applying v2 – u2 | = | 2aS |
| (0)2 – (120)2 | = | 2 × a × 1200 |
| ∴ a | = | – 14400/2400 ms–2 |
| = | – 6 ms–2. | |
| ∴ Retardation | = | – (a) |
| = | – (– 6 ms–2) | |
| = | 6 ms–2 |
| 2) Applying v | = | u + at |
| 0 | = | 120 – 6 × t |
| ∴ t | = | 120/6 s |
| = | 20s. |
A car, initially at rest, picks up a velocity of 108 km hr–1 in 40 s. Calculate its acceleration and distance covered by car?
| Given Initial velocity of car (u) | = | 0 |
| Final velocity of car (v) | = | 108 km hr–1 |
| = | 30 ms–1 | |
| Time (t) | = | 40 s. |
| 1) Applying v | = | u + at. |
| 30 | = | 0 + a × 40 |
| ∴ a | = | 30/40 ms–2 |
| = | 0.75 ms–2. | |
| 2) Applying S | = | (ut) +((1/2) × at2) |
| = | (0 × 40) + 1/2 × 0.75 × (40)2) | |
| = | 600 m. |
An aeroplane touches down at 360 km hr–1 and stops after 1.5 minutes. Calculate its acceleration and length of runway?
| Given Initial velocity of car (u) | = | 360 km hr–1 |
| = | 100 ms–1. | |
| Final velocity of car (v) | = | 0. |
| Time (t) | = | 1.5 min |
| = | 90 s. |
| 1) Applying v | = | u + at. |
| 0 | = | 100 + a × 90 |
| ∴ a | = | – 100/90 ms–2 |
| = | – 1.11 ms–2. | |
| 2) Applying S | = | [ut + ((1/2)at2)] |
| = | [(100 × 90) + ((1/2) × (–1.11) × (90)2)] | |
| = | (9000 – 4495.5) m | |
| = | 4504.5 m. |
A car running at 72 km h–1 is slowed down to 18 km h–1 by the application of brakes over a distance of 20 m. Calculate de–acceleration of car. If the brakes are applied with the same force calculate total time in which car stops and total distance covered by it?
| Case (i): | ||
| Given u | = | 72 km hr–1 |
| = | 20 ms–1 | |
| v | = | 18 km hr–1 |
| = | 5 ms–1 | |
| S | = | 20 m |
| and a | = | ? |
| Applying v2 – u2 | = | 2aS |
| (5)2 – (20)2 | = | 2 × a × 20 |
| ∴ a | = | – 375/40 |
| = | – 9.375 ms–2 | |
| ∴ De–acceleration | = | – (a) |
| = | – (– 9.375 ms–2) | |
| = | 9.375 ms–2 |
| Case (ii): | ||
| Given u | = | 72 km hr–1 |
| = | 20 ms–1 | |
| v | = | 0 ms–1 |
| and a | = | 9.375 ms–2 |
| Applying v | = | u + at |
| 0 | = | 20 – 9.375 × t |
| ∴ t | = | 2.13 s. |
| Applying S | = | [(ut + (1/2 at2))] |
| = | [(20 × 2.13 – (1/2 × 9.375 × (2.13)2)] | |
| = | 21.33 m. |
A stone dropped from the top of a tower, takes 10 seconds to reach ground level. Calculate Final velocity of stone, height of the tower and distance traveled by stone in 3rd second of its motion, (take g = 10 ms–2)?
| Given Initial velocity of stone (u) | = | 0 |
| Acceleration due to gravity(g) | = | 10 ms–2 |
| Time (t) | = | 10 s. |
| 1)Applying v | = | u + gt |
| = | 0 + 10 × 10 | |
| = | 100 ms–1 |
| 2) Applying S | = | ut + 1/2gt2 |
| = | [(0 × 10) + ((1/2)(10 × (10)2))] | |
| = | 500 m |
| 3) Applying Dnth | = | u + (a/2)(2n – 1) |
| D3rd | = | 0 + (10/2)(2 × 3 – 1) |
| = | 25 m. |
A stone projected vertically upwards, takes 3.2 seconds to reach highest point. Calculate initial velocity of stone and maximum height attained by it? (Take g = 10 ms–2)
| Given Initial velocity of stone (u) | = | ? |
| Final velocity of stone (v) | = | 0 ms–2 |
| Acceleration due to gravity(g) | = | – 10 ms–2 |
| Time (t) | = | 3.2 s. |
| 1) Applying v | = | u + gt |
| 0 | = | u – 10 × 3.2 |
| u | = | 32 ms–1 |
| 2) Applying S | = | ut + 1/2 gt2 |
| = | 32 × 3.2 + 1/2 × (–10) × (3.2)2 | |
| = | 102.4 – 51.2 | |
| = | 51.2 m. |
A stone dropped from the top of a cliff reaches the ground level after 3s and goes 50 cm deep in the sand. Calculate height of cliff, final velocity of stone and the de–acceleration produced by sand? (Take g = 9.8 ms–2)
| Case (i): | ||
| When stone falls through air, u | = | 0 ms–1, |
| t | = | 3s and g = 9.8 ms–2 |
| v | = | ? |
| and S | = | ? |
| Applying v | = | u + gt. |
| = | 0 + 9.8 × 3 | |
| = | 29.4 ms–1 | |
| Applying S | = | ut + 1/2gt2 |
| = | 0 × 3 + 1/2 × 9.8 × (3)2 | |
| = | 44.1 m. |
| Case (ii): | ||
| When stone travels in sand, | ||
| u | = | 29.4 ms–1, |
| v | = | 0 |
| and S | = | 50 cm |
| = | 0.5 m. | |
| Applying v2 – u2 | = | 2aS. |
| (0)2 – (29.4)2 | = | 2 × a × 0.5 |
| a | = | – 29.4 × (29.4/2) × 0.5 ms–2 |
| = | – 864.36 ms–2 | |
| ∴ De–acceleration | = | – (a) |
| = | – (– 864.36)ms–2 | |
| = | 864.36 ms–2 |
A vehicle covers a distance of 20 m in 5th second and 35 m in 8th second of its motion. Calculate the acceleration and initial velocity of vehicle?
| Given | ||
| D5th | = | 20m, |
| D8th | = | 35m. |
| ∴ D5th | = | u + a/2(2n – 1) |
| 20 | = | u + a/2(2 × 5 – 1) |
| (or) 40 | = | 2u + 9a ------ (i) |
| ∴ D8th | = | u + a/2(2n – 1) |
| 35 | = | u + a/2(2 × 8 – 1) |
| (or) 70 | = | 2u + 15a -----(ii) |
| Rewriting (i) and (ii) and solving simultaneously, | ||
| 2u + 9a | = | 40 |
| – 6a | = | – 30 |
| ∴ a | = | 5 ms–2 |
| Putting value of 'a' in (i) | ||
| 2u + 9 × 5 | = | 50 |
| ∴ u | = | 2.5 ms–1. |
A spaceship traveling in space at 200 km s–1, fires its engine for 10 seconds, such that its final velocity is 400 km s–1. Calculate the total distance traveled by the ship in one minute starting from the time of firing?
| Case (i): | ||
| When spaceship is accelerating: | ||
| Initial velocity (u) | = | 200 km s–1 |
| Time(t) | = | 10s. |
| Final velocity (v) | = | 400 km s–1 |
| ∴ Applying v | = | u + gt. |
| 400 | = | 200 + a × 10 |
| 10a | = | 200 |
| (or) a | = | 20 km s–1 |
| ∴ Applying S | = | ut + 1/2 at2 |
| = | 200 × 10 + (1/2 × 20 × (10)2) | |
| = | 2000 + 1000 | |
| = | 3000 km. |
| Distance covered | = | uniform velocity × time |
| = | 400 × 50 | |
| = | 20000 km | |
| ∴ Total distance traveled by spaceship in one minute. | ||
| = | 3000 + 20000 km | |
| = | 23,000 km. |
A motor bike, initially at rest, picks up a velocity of 72 km h –1 over a distance of 40 m. Calculate acceleration and time in which it picks up above velocity ?
| Initial velocity of car (u) | = | 0 |
| Final velocity of car (v) | = | 72 km hr–1 |
| = | 20 ms–1 | |
| Distance covered by car (S) | = | 40 m. |
| Acceleration (a) | = | ? |
| and Time (t) | = | ? |
| 1) Applying v2 – u2 | = | 2aS |
| (20)2 – (0)2 | = | 2 × a × 40 |
| ∴ a | = | 400/80 ms–2 |
| = | 5 ms–2 |
| 2) Applying v | = | u + at. |
| 20 | = | 0 + 5 × t |
| ∴ t | = | 20/5 s |
| = | 4 s. |
| Initial velocity of cyclist (u) | = | 5 ms–1 |
| Final velocity of cyclist (v) | = | 10 ms–1 |
| Distance covered by cyclist (S) | = | 50 m |
| Acceleration (a) | = | ? |
| and Time (t) | = | ? |
| 1) Applying v2 – u2 | = | 2aS. |
| (10)2 – (5)2 | = | 2 × a × 50 |
| ∴ a | = | 100 – 25/100 ms–2 |
| = | 75/100 ms–2 | |
| = | 0.75 ms–2 |
| 2) Applying v | = | u + at. |
| 10 | = | 5 + 0.75 × t |
| ∴ t | = | (10 – 5)/0.75 s |
| = | 6.67 s. |
An aeroplane lands at 216 km hr–1 and stops after covering a runway of 2 km.Calculate the acceleration and the time, in which it comes to rest ?
| Initial velocity of aeroplane (u) | = | 216 km hr–1 |
| = | 60 ms–1 | |
| Final velocity of aeroplane (v) | = | 0 ms–1 |
| Distance (S) | = | 2 km |
| = | 2000 m. |
| 1) Applying v2 – u2 | = | 2aS. |
| (0)2 – (60)2 | = | 2 × a × 2000 |
| ∴ a | = | – 3600/4000 ms–2 |
| = | – 0.9 ms–2 | |
| ∴ Retardation | = | – (a) |
| = | – (– 0.9 ms–2) | |
| = | 0.9 ms–2 |
| 2) Applying v | = | u + at. |
| 0 | = | 60 – 0.9 × t |
| ∴ t | = | 60/0.9 s |
| = | 66.67 s. |
A truck running at 90 km hr–1 , is brought to rest over a distance of 25 m. Calculate the retardation and time for which brakes are applied ?
| Initial velocity of truck (u) | = | 90km hr–1 |
| = | 25 ms–1 | |
| Final velocity of truck (v) | = | 0 |
| Distance (S) | = | 25 m. |
| 1)Applying v2 – u2 | = | 2aS. |
| (0)2 – (25)2 | = | 2 × a × 25 |
| ∴ a | = | –625/50 ms–2 |
| = | –12.5 ms–2 | |
| Retardation | = | –(a) |
| = | 12.5 ms–2 |
| 2) Applying v | = | u + at. |
| 0 | = | 25 + (–12.5) × t |
| ∴ t | = | –25/–12.5 s |
| = | 2 s. |
A racing car, initially at rest, picks up a velocity of 180 km hr–1 in 4.5s. Calculate acceleration and distance covered by car?
| Initial velocity of car (u) | = | 0 |
| Final velocity of car (v) | = | 180 km hr–1 |
| = | 50 ms–1 | |
| Time (t) | = | 4.5 s. |
| 1)Applying v | = | u + at. |
| 50 | = | 0 + a × 4.5 |
| ∴ a | = | 50/4.5 ms–2 |
| = | 11.11 ms–2 | |
| 2) Applying S | = | ut + 1/2 at2 |
| = | (0 × 4.5) + 1/2 × 11.11 × (4.5)2 | |
| = | 112.5 m. |
A motor bike running at 5 ms–1, picks up a velocity of 30 ms–1 in 5s. Calculate acceleration and distance covered during acceleration?
| Initial velocity of motor bike (u) | = | 5 ms–1 |
| Final velocity of motor bike (v) | = | 30 ms–1 |
| Time (t) | = | 5 s. |
| 1) Applying v | = | u + at. |
| 30 | = | 5 + a × 5 |
| ∴ a | = | (30 – 5)/5 ms–2 |
| = | 5 ms–2 | |
| 2) Applying S | = | ut + 1/2 at2 |
| = | [(5 × 5) + ((1/2) × 5 × (5)2)] | |
| = | 87.5 m. |
A motor bike running at 90 km hr–1 is slowed down to 18 km hr–1 in 2.5 s. Calculate acceleration and distance covered during slow down?
| Initial velocity of motor bike (u) | = | 90 km hr–1 |
| = | 25 ms–1 | |
| Final velocity of motor bike (v) | = | 18 km hr–1 |
| = | 5 ms–1 | |
| Time (t) | = | 2.5 s. |
| 1) Applying v | = | u + at. |
| 5 | = | 25 + a × 2.5 |
| ∴ a | = | (5 – 25)/2.5 ms–2 |
| = | – 8 ms–2 | |
| 2) Applying S | = | ut + 1/2 at2 |
| = | [(25 × 2.5) + ((1/2) × (–8) × (2.5)2)] | |
| = | (62.5 – 25) m | |
| = | 37.5 m. |
A cyclist driving at 36 km hr–1 stops his mount in 2s, by the application of brakes. Calculate retardation and distance covered during the application of brakes?
| Initial velocity of cyclist (u) | = | 36 km hr–1 |
| = | 10 ms–1 | |
| Final velocity of cyclist (v) | = | 0. |
| Time (t) | = | 1.5 min |
| = | 2 s. |
| 1) Applying v | = | u + at. |
| 0 | = | 10 + a × 2 |
| ∴ a | = | –10/2 ms–2 |
| = | – 5 ms–2 | |
| Retardation | = | – (a) |
| = | 5 ms–2 | |
| 2) Applying S | = | ut + 1/2 at2 |
| = | [(10 × 2) + ((1/2)× (–5) × (2)2)] | |
| = | (20 – 10) m | |
| = | 10 m. |
A motor bike running at 90 km hr–1, is slowed down to 54 km hr–1 by the application of brakes, over a distance of 40 m. If the brakes are applied with the same force, calculate total time in which bike comes to rest and total distance traveled by bike?
| Given u | = | 90 km hr–1 |
| = | 25 ms–1 | |
| v | = | 54 km hr–1 |
| = | 15 ms–1 | |
| S | = | 40 m |
| and a | = | ? |
| Applying v2 – u2 | = | 2aS. |
| (15)2 – (25)2 | = | 2 × a × 40 |
| ∴ a | = | – 400/80 |
| = | – 5 ms–2 | |
| ∴ De–acceleration | = | – (a) |
| = | – (– 5 ms–2) | |
| = | 5 ms–2 |
| Case (i): | ||||
| Given u | = | 90 km hr–1 = 25 ms–1 | ||
| v | = | 0 ms–1 | ||
| and a | = | 5 ms–2 | ||
| Applying v | = | u + at | ||
| 0 | = | 25 – 5 × t | ||
| ∴ t | = | 5 s. | ||
| Case (ii): | ||||
| Applying S | = | ut + 1/2 at2 | ||
| = | 25 × 5 – 1/2 × 5 × (5)2 | |||
| = | 62.5 m. |
A motor car slows down from 72 km hr–1 to 36 km hr–1 over at distance of 25 m. If the brakes are applied with the same force calculate total time in which car comes to rest and distance traveled by it?
| Given u | = | 72 km hr–1 |
| = | 20 ms–1 | |
| v | = | 36 km h–1 |
| = | 10 ms–1 | |
| S | = | 25 m |
| and a | = | ? |
| Applying v2 – u2 | = | 2aS |
| (10)2 – (20)2 | = | 2 × a × 25 |
| ∴ a | = | – 300/50 |
| = | – 6 ms–2 | |
| ∴ De–acceleration | = | – (a) |
| = | – (– 6 ms–2) | |
| = | 6 ms–2 | |
| Case (i): | ||
| Given u | = | 72 km hr–1 |
| = | 20 ms–1 | |
| v | = | 0 ms–1 |
| and a | = | – 6 ms–2 |
| Applying v | = | u + at |
| 0 | = | 20 – 6 × t |
| ∴ t | = | 3.33 s. |
| Case (ii): | ||
| Applying S | = | ut + 1/2 at2 |
| = | 20 × 3.33 – 1/2 × 6 × (3.33)2 | |
| = | 33.33 m. |