Complete the following table: Car Model Driver X Reaction time 0 . 20 s Driver Y Reaction time 0 . 30 s A (deceleration on hard-braking = 6 . 0 m s - 2 ) Speed = 54 km h - 1 Braking distance a = ............ Total stopping distance b = ............ Speed = 72 km h - 1 Braking distance c = ........... Total stopping distance d = ............ B (deceleration on hard-braking = 7 . 5 m s - 2 ) Speed = 54 km h - 1 Braking distance e = ........... Total stopping distance f = ............ Speed = 72 km h - 1 Braking distance g = ............. Total stopping distance h = ............

Distance after which the car stops after brakes were applied is called braking distance. But due to reaction time of driver, he always travels some extra distance, irrespective of the car model Braking distance can be calculated through third equation of motion for all drivers for each case. v 2 = u 2 + 2 a s ⇒ s = - u 2 2 a ∵ v = 0 For driver X : Initial velocity, u = 54 km h - 1 = 54 × 5 18 = 15 m s - 1 . Final velocity, v = 0 , Before braking, distance traveled by X = speed × time = 15 × 0 . 20 = 3 m Braking distance for car A = ( 15 ) 2 - 2 × - 6 = 18 . 75 m ≈ 19 m . Braking distance for car B = ( 15 ) 2 - 2 × - 7 . 5 = 15 m . Total stopping distance for A = 3 + 19 = 22 m . Total stopping distance for B = 3 + 15 = 18 m . For driver Y : Initial velocity, u = 72 km h - 1 = 72 × 5 18 = 20 m s - 1 . Final velocity, v = 0 . Before braking, distance traveled by Y = speed × time = 20 × 0 . 30 = 6 m Braking distance for car A = 20 2 2 × 6 = 33 . 3 ≈ 33 m . Braking distance for car B = 20 2 2 × 7 . 5 = 26 . 7 ≈ 27 m . Total stopping distance for A = 6 + 33 = 39 m . Total stopping distance for B = 6 + 27 = 33 m . Car Model Driver X Reaction Time = 0 . 20 s Driver Y Reaction Time = 0 . 30 s A (deceleration on hard-braking = 6 . 0 m s - 2 ) Speed = 54 km h - 1 Braking distance , a = 19 m Total stopping distance, b = 22 m Speed = 72 km h - 1 Braking distance, c = 33 m Total stopping distance, d = 39 m B (deceleration on hard braking = 7 . 5 m s - 2 ) Speed = 54 km h - 1 Breaking distance, e = 15 m Total stopping distance, f = 18 m Speed = 72 km h - 1 Braking distance, g = 27 m Total stopping distance, h = 33 m

MEDIUM

JEE Main

IMPORTANT

Earn 100

A driver takes $0.20 s$ to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of $54 km h^{- 1}$ and the brakes cause a deceleration of $6.0 m s^{- 2}$ , find the distance traveled by the car after he sees the need to put the brakes.

Important Questions on Rest and Motion : Kinematics

MEDIUM

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 20

Complete the following table:

Car Model	Driver $X$ Reaction time $0.20 s$	Driver $Y$ Reaction time $0.30 s$
$A$ (deceleration on hard-braking $= 6.0 m s^{- 2}$ )	Speed $= 54 km h^{- 1}$ Braking distance $a =$ ............ Total stopping distance $b =$ ............	Speed $= 72 km h^{- 1}$ Braking distance $c =$ ........... Total stopping distance $d =$ ............
$B$ (deceleration on hard-braking $= 7.5 m s^{- 2}$ )	Speed $= 54 km h^{- 1}$ Braking distance $e =$ ........... Total stopping distance $f =$ ............	Speed $= 72 km h^{- 1}$ Braking distance $g =$ ............. Total stopping distance $h =$ ............

MEDIUM

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 21

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turn at a speed of

72 km h^{- 1}

. The jeep follows it at a speed of

90 km h^{- 1}

, crossing the turn ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turn will the jeep catch up with the bike?

MEDIUM

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 22

A car travelling at

60 km h^{- 1}

overtakes another car travelling at

42 km h^{- 1}

. Assuming each car to be

5.0 m

long, find the time taken during overtake and the total road distance used for overtake.

EASY

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 23

A ball is projected vertically upwards with a speed of

50 m s^{- 1}

. Find the maximum height.

EASY

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 23

A ball is projected vertically upwards with a speed of

50 m s^{- 1}

. Find the time required to reach the maximum height.

EASY

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 23

A ball is projected vertically upwards with a speed of

50 m s^{- 1}

. Find the speed at half the maximum height. (Take

g = 10 m s^{- 2}

MEDIUM

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 24

A ball is dropped from a balloon going up at a speed of

7 m s^{- 1}

. If the balloon was at a height

60 m

at the time of dropping the ball, how long will the ball take in reaching the ground? (

g = 10 m s^{- 2}

)

EASY

JEE Main

IMPORTANT

CONCEPTS OF PHYSICS [VOLUME 1]>Chapter 3 - Rest and Motion : Kinematics>EXERCISES>Q 25

A stone is thrown vertically upwards with a speed of

28 m s^{- 1}

. Find the maximum height reached by the stone.

A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km h-1 and the brakes cause a deceleration of 6.0 m s-2, find the distance traveled by the car after he sees the need to put the brakes.

Important Questions on Rest and Motion : Kinematics

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