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

Index

## 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 vUE= 0ms-1

Distance S = 1km = 1000m

time = 100s

top speed vF=?

S = VUEx 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 1calleequation of motion

vF= VUE+ em

vF= 0 + 0,2 x 100

vF= 20ms-1

Announcement

#### 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 = Vue =10ms-1

Acceleration a = 0.2ms-2

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

• distance S =?
• top speed vF=?

S = VUEx 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 1calleequation of motion

vF= VUE+ em

vF= 10 + 0,2 x 30

vF= 10 + 6

vF= 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 = Vue =30ms-1

Acceleration due to gravity g = -10 ms-2

Time to reach maximum altitude = t = 3 s

top speed vF= 0ms-1

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

S = VUEx 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

#### (-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 =?

vF= VUE+ em

o and = VF-vUE

one = VF-vUE/T

a = 0 - 40/10

a = -4ms-2

Total distance traveled = S = S1+S2

using this relationship

S1= VT

S1= 40x5

S1= 200m ………………………………. (I)

now with 3thirdequation of motion

2aS2= VF2–vUE2

S2= VF2–vUE2/2a

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

Announcement

#### 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 vUE= 0ms-1

Acceleration a = 0.5ms-2

Distance S = 100 m

top speed vF=?

2aS = Vf2–vUE2

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;

vF= 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 = VUE= 0ms-1

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

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

S1= VVonxt

S1= (VF+vUE)/2 x t

S1= 13.333 + 0/2 x 120

S1= 6,6665 x 120

S1= 799,99m = 800m

Autumn – II:

Uniform speed = VF= 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 = VF= 13.333ms-1

Final speed = VUE= 0ms-1

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

S3= VVonxt

S3= (VF+vUE)/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

Announcement

#### (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 = VF= 0ms-1

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

without youF= VUE+ gxt

vUE= VF- gxt

vUE= 0 – (-10) x 3

vUE= 30ms-1

now with 3thirdequation of motion

2aS = Vf2–vi2

S =vf2–vi2/2a

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 = Vue =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 = VF2–vUE2

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:

vF= 0, S2=?

2aS = VF2–vUE2

-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

Announcement

#### 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 = VUE= 96 km/h-1= 96 x 1000/3600 = 96000/3600ms-1

Final speed = VF= 0ms-1

a = -(48000/3600)2x 3/1600ms-2

time = t =?

you vF= VUE+ em

o and = VF–vUE

t = VF–vUE/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

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