Kinematics - Numerical Problems - Lecture 9 Physics - Lecture notes (2023)

numerical problems

2.1 A train moves at a constant speed of 36 km/h-1 for 10 s. Calculate the distance traveled. (100m)

Solution:Speed ​​= V = 36 km/h^-1= 36 x 1000/ 60 x 60 = 36000/ 3600 = 10ms^-1

time t = 10s

distance = S =?

S = Vt

P = 10 x 10 = 100 m

2.2 A train starts from rest. It covers 1 km in 100 s with constant acceleration. What will its velocity be after 100 s? (20ms^-1)

Solution:initial velocity v_UE= 0ms^-1

Distance S = 1km = 1000m

time = 100s

top speed v_F=?

S = V_UEx t + 1/2 x a x t2

1000 = 0 x 100 + 1/2 x a x (100)²

1000 = 1/2 x 10000a

1000 = 5000a

a = 1000/5000 = 0,2ms^-2

now with 1^calleequation of motion

v_F= V_UE+ em

v_F= 0 + 0,2 x 100

v_F= 20ms^-1

2.3 A car has a speed of 10 ms^-1. Speed ​​up in 0.2ms^-2for half a minute. Find the distance traveled in this time and the final speed of the car. (390m, 16ms^-1)

Solution:Initial speed = V_{ue =}10ms^-1

Acceleration a = 0.2ms^-2

Time t = 0.5 min = 0.5 x 60 = 30 s

• distance S =?
• top speed v_F=?

S = V_UEx t + 1/2 x a x t²

S= 10 x 30 + 1/2 x 0,2 x (30)²

S = 300 + 1/2 x 0,2 x 90

S = 300 + 1/2 x 2/10 x 90

S = 300 + 90

S = 390 m

• use of 1^calleequation of motion

v_F= V_UE+ em

v_F= 10 + 0,2 x 30

v_F= 10 + 6

v_F= 16ms^-1

2.4 A tennis ball is thrown vertically upward with a velocity of 30 ms.^-1, it takes 3 s to reach the highest point. Calculate the maximum height the ball will reach. How long will it be until I'm back on the ground? (45m, 6s)

Solution:Initial speed = V_{ue =}30ms^-1

Acceleration due to gravity g = -10 ms^-2

Time to reach maximum altitude = t = 3 s

top speed v_F= 0ms^-1

(UE)Maximum height reached by the sphere S =?

(II)Time to return to the ground t =?

S = V_UEx t + 1/2 x g x t²

S = 30 x 3 + 1/2 x (-10) x (3)²

P = 90 – 5 x 9

S = 90 – 45

P = 45m

Total time = time to reach maximum altitude + time to return to the ground

= 3s + 3s = 6s

2.5 A car moves with a uniform speed of 40 ms^-1for 5 sec. It stops for the next 10 s with a uniform delay.

Meet:

• delay

• total distance traveled by the car.

(-4ms^-2, 400m)

Solution:Initial velocity = Vi = 40 ms^-1

time = t = 5s

Final speed = Vf = 0 ms^-1

time = 10s

• delay a =?
• Total distance S =?

v_F= V_UE+ em

o and = V_F-v_UE

one = V_F-v_UE/T

a = 0 - 40/10

a = -4ms^-2

Total distance traveled = S = S₁+S₂

using this relationship

S₁= V_T

S₁= 40x5

S₁= 200m ………………………………. (I)

now with 3^thirdequation of motion

2aS₂= V_F²–v_UE²

S₂= V_F²–v_UE²/2a

S₂= (0)2 – (40)2/2 x (-4)

S₂= -1600/-8

S₂= 200 m ………………………………………… (ii)

From (i) and (ii) we get;

S = S₁+S₂

Oder S = 200 m + 200 m

S = 400 m

2.6 A train starts from a standstill with an acceleration of 0.5 ms^-2. Find his speed in kmh-1 when he has moved 100m. (36 km/h^-1)

Solution:initial velocity v_UE= 0ms^-1

Acceleration a = 0.5ms^-2

Distance S = 100 m

top speed v_F=?

2aS = Vf²–v_UE²

2 x 0,5 x 100 = Vf²– 0

You, 100 = Vf²

you vf²= 100ms^-1……………………..(UE)

speed in km/h^-1:

From (I) we obtain;

v_F= 10 x 3600/1000 = 36 km/h^-1

2.7 A train starts from rest, accelerates constantly and reaches a speed of 48 km/h^-1in 2 minutes At that speed, he drives for 5 minutes. Finally, it moves with a uniform delay, stopping after 3 minutes. Calculate the total distance traveled by the train.

Solution: Case – I: (6000 m)

Initial speed = V_UE= 0ms^-1

Time = t = 2 minutes = 2x 60 = 120 s

Final speed = V_F= 48 km/h^-1= 48 x 1000/3600 = 13 333 ms^-1

S₁= V_Vonxt

S₁= (V_F+v_UE)/2 x t

S₁= 13.333 + 0/2 x 120

S₁= 6,6665 x 120

S₁= 799,99m = 800m

Autumn – II:

Uniform speed = V_F= 13.333ms^-1

Time = t = 5 minutes = 5 x 60 = 300 s

S₂= v x t

S₂= 13.333 x 300

S₂= 3999,9 = 4000m

Autumn – III:

Initial speed = V_F= 13.333ms^-1

Final speed = V_UE= 0ms^-1

Time = t = 3 minutes = 3 x 60 = 180 s

S₃= V_Vonxt

S₃= (V_F+v_UE)/2x180

S₃= 13.333 + 0/2 x 180

S₃= 6,6665 x 180

S₃= 1199,97 = 1200m

Total distance = S = S₁+S₂+S₃

S = 800 + 4000 + 1200

S = 6000 m

2.8 A cricket ball is kicked upwards and 6 s later it falls back to the ground. Calculation

(i) Maximum height reached by the ball

(ii) Initial velocity of the ball (45m, 30ms-1)

Solution:Acceleration due to gravity = g = -10 ms^-1(for upward movement)

Time to reach maximum altitude (one-sided time) = t = 6/2 = 3 s

Speed ​​at maximum altitude = V_F= 0ms^-1

• Maximum height reached by the sphere S = h =?
• The maximum initial velocity of the ball = V_UE=?

without you_F= V_UE+ gxt

v_UE= V_F- gxt

v_UE= 0 – (-10) x 3

v_UE= 30ms^-1

now with 3^thirdequation of motion

2aS = V_f2–v_i2

S =v_f2–v_i2/2a

S = (0)2 – (30)2/2 x (-10)

S = -90/-20

P = 45m

2.9 When braking, the speed of a train decreases from 96 km/h-1 to 48 km/h-1 in 800 m. How far does the train travel before stopping? (Assuming the delay is constant.) (266.66m)

Solution:Initial speed = V_{ue =}96 km/h^-1= 96 x 1000/3600 = 96000/3600ms^-1

Final speed = Vf = 48 km/h^-1= 48 x 1000/3600 = 48000/3600ms^-1

Distance = S = 800 m

More distance = S₁=?

First we find the value of the acceleration a

2aS = V_F²–v_UE²

2 x a x 800 = (48000/3600)²– (96000/3600)²

1600a = (48000/3600)²– ((2x48000)/3600)²(96000 = 2x48000)

1600a = (48000/3600)²((1)²– (2)²) (with (48000/36000) as usual)

1600a = (48000/3600)²(1 – 4)

1600a = (48000/3600)²(-3)

= (48000/3600)²×3/1600

now we find the value of the additional distance S²:

v_F= 0, S₂=?

2aS = V_F²–v_UE²

-2 (48000/36000)²x 3/1600 x S₁= (0)²– (48000/36000)²

S₁= (48000/36000)²x (48000/36000)²x 1600/3 x 2

S₁= 1600/6

S₂= 266,66m

2.10 In the problem above, find the time it takes for the train to come to a stop after braking. (80s)

Solution:taking the data from exercise 2.9:

Initial speed = V_UE= 96 km/h^-1= 96 x 1000/3600 = 96000/3600ms^-1

Final speed = V_F= 0ms^-1

a = -(48000/3600)²x 3/1600ms^-2

time = t =?

you v_F= V_UE+ em

o and = V_F–v_UE

t = V_F–v_UE/A

t = 0 – (48000/3600)/-( (48000/3600) x 1/1600

t = – 96000/3600 x (3600/48000)²×1600/3

t = 2 x 48000/3600 x (3600/48000 x 3600/48000) x 1600/3

t = 2x3600/3x3

t = 2 x 40 = 80 s