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})

^{-1})

**Solution:**initial velocity v_{UE}= 0ms^{-1}

Distance S = 1km = 1000m

time = 100s

top speed v_{F}=?

S = V_{UE}x 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^{calle}equation of motion

v_{F}= V_{UE}+ em

v_{F}= 0 + 0,2 x 100

v_{F}= 20ms^{-1}

Announcement

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

^{-1}. Speed up in 0.2ms

^{-2}for 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_{UE}x t + 1/2 x a x t^{2}

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
^{calle}equation 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)

^{-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_{UE}x t + 1/2 x g x t^{2}

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**^{-1}for 5 sec. It stops for the next 10 s with a uniform delay.

^{-1}for 5 sec. It stops for the next 10 s with a uniform delay.

**Meet:**

**delay****total distance traveled by the car.**

**(-4ms**^{-2}, 400m)

^{-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 = S_{1}+S_{2}

using this relationship

S_{1}= V_{T}

S_{1}= 40x5

S_{1}= 200m ………………………………. (I)

now with 3^{third}equation of motion

2aS_{2}= V_{F}^{2}–v_{UE}^{2}

S_{2}= V_{F}^{2}–v_{UE}^{2}/2a

S_{2}= (0)2 – (40)2/2 x (-4)

S_{2}= -1600/-8

S_{2}= 200 m ………………………………………… (ii)

From (i) and (ii) we get;

S = S_{1}+S_{2}

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})

^{-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 = Vf^{2}–v_{UE}^{2}

^{ }2 x 0,5 x 100 = Vf^{2}– 0

You, 100 = Vf^{2}

^{ }you vf^{2}= 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**^{-1}in 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.

^{-1}in 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}

S_{1}= V_{Von}xt

S_{1}= (V_{F}+v_{UE})/2 x t

S_{1}= 13.333 + 0/2 x 120

S_{1}= 6,6665 x 120

S_{1}= 799,99m = 800m

**Autumn – II:**

Uniform speed = V_{F}= 13.333ms^{-1}

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

S_{2}= v x t

S_{2}= 13.333 x 300

S_{2}= 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

S_{3}= V_{Von}xt

S_{3}= (V_{F}+v_{UE})/2x180

S_{3}= 13.333 + 0/2 x 180

S_{3}= 6,6665 x 180

S_{3}= 1199,97 = 1200m

Total distance = S = S_{1}+S_{2}+S_{3}

S = 800 + 4000 + 1200

S = 6000 m

Announcement

**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^{third}equation 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 = S_{1}=?

First we find the value of the acceleration a

2aS = V_{F}^{2}–v_{UE}^{2}

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 S^{2}:

v_{F}= 0, S_{2}=?

2aS = V_{F}^{2}–v_{UE}^{2}

-2 (48000/36000)^{2}x 3/1600 x S_{1}= (0)^{2}– (48000/36000)^{2}

S_{1}= (48000/36000)^{2}x (48000/36000)^{2}x 1600/3 x 2

S_{1}= 1600/6

S_{2}= 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 = V_{UE}= 96 km/h^{-1}= 96 x 1000/3600 = 96000/3600ms^{-1}

Final speed = V_{F}= 0ms^{-1}

a = -(48000/3600)^{2}x 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