A train moves with a uniform velocity of 36 $kmh^{-1} for 10s$. Find the distance traveled by it.                                                                                                  (100 m)

Difficulty: Easy

Solution:

Velocity = V = $36 kmh^{-1}$ = $36 \times \frac{1000}{60 \times 60}$ =$\frac{36000}{3600}$ = $10 ms^{-1}$
Time t = 10 s
Distance = S =?
S = Vt
S = $10 \times 10$ = 100 m

A train starts from rest. It moves through 1 km in 100 s with uniform acceleration. What will be its speed at the end of 100 s.?   $(20 ms^{-1})$

Difficulty: Easy

Solution:

Initial velocity Vi = 0 $ms^{-1}$
Distance S = 1 km = 1000 m
Time = 100 s
Final velocity Vf =?
S = $Vi \times t +\frac{1}{2} \times a \times t2$
1000 = $0 \times100 + \frac{1}{2} \times a \times (100)^{2}$
1000 = $\frac{1}{2} \times 10000a$
1000 = 5000a
a = 1000/5000 = $0.2 ms^{-2}$

\Now using 1st equation of motion

Vf = Vi + at
Vf = $0 + 0.2 \times 100$
Vf = $20 ms^{-1}$

A car has a velocity of $10 ms^{-1}$. It accelerates at $0.2 ms^{-2}$ for half a minute. Find the distance traveled during this time and the final velocity of the car. $(390 m, 16 ms^{-1})$

Difficulty: Easy

Solution:

Initial velocity = Vi = $10 ms^{-1}$
Acceleration a = 0.2 $ms^{-2}$
Time t = 0.5 min = $0.5 \times60$ = 30 s

(I) Distance S =?
(II) Final velocity Vf =?

S = $Vi \times t + \frac{1}{2} \times a \times t2$
S=  $10 \times 30 + \frac{1}{2} \times 0.2 \times (30)^{2}$
S = $300 + \frac{1}{2}\times 0.2 \times 90$
S = $300 + \frac{1}{2} \times \frac{2}{10} \times$ 90
S = 300 + 90
S = 390 m

(II)Using 1st equation of motion
Vf = Vi + at
Vf = $10 + 0.2 \times30$
Vf = 10 + 6
Vf = $16 ms^{-1}$

A tennis ball is hit vertically upward with a velocity of 30 $ms^{-1}$, it takes 3 s to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return to the ground? (45 m, 6 s)

Difficulty: Easy

Solution:

Initial velocity = Vi = $30 ms^{-1}$
Acceleration due to gravity g = $-10 ms^{-2}$
Time to reach maximum height = t = 3 s
Final velocity Vf = $0 ms^{-1}$

(I)Maximum height attained by the ball S =?

S = $Vi \times t + \frac{1}{2} \times g \times t2$
S = $30 \times 3 + \frac{1}{2}\times (-10) \times (3)^{2}$
S = $90 – 5 \times 9$
S = 90 – 45
S = 45 m

Total time = time to reach maximum height + time to return to the ground
= 3 s + 3 s = 6 s

A car moves with a uniform velocity of 40 $ms^{-1}$ for 5 s. It comes to rest in the next 10 s with uniform deceleration.

Find:

1. Deceleration
2. Total distance traveled by car. $(-4 ms^{-2}, 400 m)$
Difficulty: Easy

Solution:

Initial velocity = Vi = 40 $ms^{-1}$
Time = t = 5 s
Final velocity = Vf = 0 $ms^{-1}$
Time = 10 s

(I) deceleration a =?
(II)total distance S =?

Vf = Vi + at
Or   at = Vf - Vi
a = $\frac{Vf - Vi}{T}$
a = $\frac{0 – 40}{10}$
a = $-4 ms^{-2}$
Total distance travelled = S = S1 + S2
By using this relation
S1 = Vt
S1 = $40 \times 5$
S1 = 200 m ……………………. (i)

Now by using 3rd equation of motion
2aS_${2} = Vf^{2} – Vi^{2}$
S2 = $Vf^{2} –\frac{Vi2}{2a}$
S2 = $(0)2 – (40)2/2\times (-4)$
S2 =$\frac{-1600}{-8}$
S2 = 200 m …………………………… (ii)

From (i) and (ii) we get;

S = S1 + S2
Or             S = 200 m + 200 m
S = 400 m

A train starts from rest with an acceleration of 0.5 $ms^{-2}$. Find its speed in $kmh^{-1}$, when it has moved through 100 m. $(36kmh^{-1})$

Difficulty: Easy

Solution:

Initial velocity Vi = 0 $ms^{-1}$
Acceleration a = $0.5 ms^{-2}$
Distance S = 100 m
Final velocity Vf =?
2aS = $Vf^{2} – Vi^{2}$
$2 \times 0.5 \times 100 = Vf^{2}$ – 0
Or         100 = $Vf^{2}$

Or

$Vf^{2} = 100 ms^{-1}$……………………..(I)
Speed in $kmh^{-1}$:
From (I) we get;
Vf = $10 \times \frac{3600}{1000}= 36 kmh^{-1}$

A train starting from rest accelerates uniformly and attains a velocity of 48 $kmh^{-1}$ in 2 minutes. It travels at this speed for 5 minutes. Finally, it moves with uniform retardation and is stopped after 3 minutes. Find the total distance traveled by train.

Difficulty: Easy

Solution:

Case – I:      (6000 m)

Initial velocity = Vi = $0ms^{-1}$
Time = t = 2 minutes = $2\times 60$ = 120 s
Final velocity = Vf = $48 kmh^{-1}$ = $48 \times \frac{1000}{3600}$ = 13.333 $ms^{-1}$
S1 = $Vav \times t$
S1 =$\frac{(Vf + Vi)}{2}\times$ t
S1 = $13.333 + \frac{0}{2} \times$120
S1 = $6.6665 \times 120$
S1 = 799.99 m = 800 m

Case – II:

Uniform velocity = Vf = $13.333 ms^{-1}$
Time = t = 5 minutes = $5 \times 60$ = 300 s
S2 = $v \times t$
S2 = $13.333 \times$ 300
S2 = 3999.9 = 4000 m

Case – III:

Initial velocity = Vf = $13.333 ms^{-1}$
Final velocity = Vi = 0 $ms^{-1}$
Time = t = 3 minutes = $3\times 60$ = 180 s
S3 = $Vav \times$ t
S3 =$\frac{(Vf + Vi)}{2}\times$ 180
S3 = $.333 +\frac{0}{2}\times$ 180
S3 = $6.6665 \times180$
S3 = 1199.97 = 1200 m
Total distance = S = S1 + S2 + S3
S = 800 + 4000 + 1200
S = 6000 m

A cricket ball is hit vertically upwards and returns to the ground 6 s later. Calculate

(i)Maximum height reached by the ball
(ii)initial velocity of the ball (45m, 30 $ms^{-1}$)

Difficulty: Medium

Solution:

Acceleration due to gravity = g = $-10 ms^{-1}$ (for upward motion)
Time to reach maximum height (one sided time) = t =$\frac{6}{2}$= 3 s
Velocity at maximum height = Vf = 0 $ms^{-1}$

(i)Maximum height reached by the ball S = h =?
(ii)The maximum initial velocity of the ball = $V_{i} =?$

Since,      $V_{f} = V_{i} + g\times t$
$V_{i} = V_{f} – g\times t$
$V_{i} = 0 – (-10) \times 3$
$V_{i}= 30 ms^{-1}$

Now using 3rd equation of motion

2aS = $V_{f}^{2} – V_{i}^{2}$
S = $V_{f}^{2}–\frac{V_{i}^{2}}{2a}$
S = (0)2 – $(30)\frac{2}{2}\times(-10)$
S = $\frac{-90}{-20}$
S = 45 m

When brakes are applied, the speed of a train decreases from 96 $kmh^{-1}$ to 48 $kmh^{-1}$ in 800 m. How much further will the train move before coming to rest? (Assuming the retardation to be constant). (266.66 m)

Difficulty: Medium

Solution:

Initial velocity = $v_{i} = 96 kmh^{-1}$ = $96\times\frac{1000}{1000}$ = $\frac{96000}{96000} ms^{-1}$
Final velocity = $v_{f} = 48 kmh^{-1}$ = 48\times $\frac{1000}{3600}$ = $\frac{48000}{3600}ms^{-1}$
Distance = S = 800 m
Further Distance = S1 =?

First of all, we will find the value of acceleration a

$2aS = v_{f}^{2} – v_{i}^{2}$
$2 \times a \times800$ = $(\frac{48000}{3600})^{2}$ – $(\frac{96000}{3600})^{2}$
1600a = $(\frac{48000}{3600})^{2}$ – $((2 \times \frac{48000}{3600})^{2}$        $(96000 = 2 \times 48000)$
1600a = $(\frac{48000}{3600})^{2} ((1)^{2} – (2)^{2})$        (taking $(\frac{48000}{36000})$ as common)
1600a = $(\frac{48000}{3600})^{2} (1 – 4)$
1600a = $(\frac{48000}{3600})^{2} (-3)$
a = $(\frac{48000}{3600})^{2} \times \frac{3}{1600 }$

Now, we will find the value of further distance S2:
$V_{f}$ = 0              ,           S2 =?
2aS = $v_{f}^{2} – v_{i}^{2}$
$-2 (\frac{48000}{36000})^{2}$ \times $\frac{3}{1600}\times$ S1 = $(0)^{2}$- $(\frac{48000}{36000})^{2}$
S1 = $(\frac{48000}{36000})^{2}$ \times $(\frac{48000}{36000})^{2}$ \times $\frac{1600}{3}\times 2$
S1 = $\frac{1600}{6}$
S2 = 266.66m

In the above problem, find the time taken by the train to stop after the application of brakes. (80 s)

Difficulty: Medium

Solution:

by taking the data from problem 2.9:
Initial velocity = $V_{i} = 96 kmh^{-1}$ = $96 \times \frac{1000}{3600}$ =$\frac{96000}{3600} ms^{-1}$
Final velocity = $V_{f} = 0 ms^{-1}$
a =$(\frac{-48000}{3600})^{2}$ \times $\frac{3}{1600} ms^{-2}$
time = t =?

Or   $V_{f}$ = $V_{i}$ + at
Or   at = $V_{f} – V_{i}$

t = $V_{f} – V_\frac{_{i}}{a}$
t = $0 \frac{-48000}{3600}$-$(\frac{48000}{3600})$ \times $\frac{1}{1600}$
t = $\frac{-96000}{3600}\times$ $(\frac{3600}{48000})^{2}$ \times  $\frac{1600}{3}$
t = $2\times \frac{48000}{3600}$\times $(\frac{3600}{48000}$\times $\frac{3600}{48000})$ \times $\frac{1600}{3}$
t = $2 \times \frac{3600}{3} \times 3$
t = $2\times 40$ = 80 s