Ch:2-Kinematics (Numerical Problems)

Table of Contents

• A train moves with a uniform velocity of 36 $kmh^{-1} for 10s$. Find the distance traveled by it.
• A train starts from rest. It moves through 1 km in 100 s with uniform acceleration.
• Find the distance traveled during this time and the final velocity of the car
• A tennis ball is hit vertically upward with a velocity of 30 $ms^{-1}$, it takes 3 s to reach the highest point.
• 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.
• A train starts from rest with an acceleration
• A train starting from rest accelerates uniformly and attains a velocity
• A cricket ball is hit vertically upwards and returns to the ground 6 s later. Calculate
• When brakes are applied, the speed of a train decreases
• In the above problem, find the time taken by the train to stop after the application of brakes

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

Difficulty: Easy

Solution:

Velocity = V = $36��{ℎ}^{-1}$ = $36\times \frac{1000}{60\times 60}$ =$\frac{36000}{3600}$ = $10�{�}^{-1}$
                       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.?   $(20�{�}^{-1})$

Difficulty: Easy

Solution:

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

\Now using 1st equation of motion 

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

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

Difficulty: Easy

Solution:           

Initial velocity = Vi = $10�{�}^{-1}$ 

Acceleration a = 0.2 $�{�}^{-2}$

Time t = 0.5 min = $0.5\times 60$ = 30 s 

 

(I) Distance S =?

(II) Final velocity Vf =?

 

       S = $��\times �+\frac{1}{2}\times �\times �2$

       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 = $10+0.2\times 30$ 

Vf = 10 + 6

Vf = $16�{�}^{-1}$

A tennis ball is hit vertically upward with a velocity of 30 

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

(I)Maximum height attained by the ball S =?
(II)Time taken to return to ground t =?

S = $��\times �+\frac{1}{2}\times �\times �2$
S = $30\times 3+\frac{1}{2}\times (-10)\times (3{)}^2$
S = $90\text{–}5\times 9$
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 $�{�}^{-1}$ for 5 s. It comes to rest in the next 10 s with uniform deceleration. 

Find:

1. Deceleration
2. Total distance traveled by car. $(-4�{�}^{-2},400�)$

Difficulty: Easy

Solution:

Initial velocity = Vi = 40 $�{�}^{-1}$ 

Time = t = 5 s

Final velocity = Vf = 0 $�{�}^{-1}$

Time = 10 s       

                                     

(I) deceleration a =?   

(II)total distance S =?     

 

Vf = Vi + at 

Or   at = Vf - Vi

                           a = $\frac{��-��}{�}$

                           a = $\frac{0\text{–}40}{10}$

                           a = $-4�{�}^{-2}$

                    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_$2=�{�}^2\text{–}�{�}^2$

                              S2 = $�{�}^2\text{–}\frac{��2}{2�}$  

                              S2 = $(0)2\text{–}(40)2\mathrm{/}2\times (-4)$

                              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 $�{�}^{-2}$. Find its speed in $��{ℎ}^{-1}$, when it has moved through 100 m. $(36��{ℎ}^{-1})$

Difficulty: Easy

Solution:         

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

Or  

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

A train starting from rest accelerates uniformly and attains a velocity of 48 

$��{ℎ}^{-1}$ 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 = $0�{�}^{-1}$ 

Time = t = 2 minutes = $2\times 60$ = 120 s

Final velocity = Vf = $48��{ℎ}^{-1}$ = $48\times \frac{1000}{3600}$ = 13.333 $�{�}^{-1}$ 

S1 = $���\times �$ 

S1 =$\frac{(��+��)}{2}\times$ 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 = $13.333�{�}^{-1}$

Time = t = 5 minutes = $5\times 60$ = 300 s 

S2 = $�\times �$ 

S2 = $13.333\times$ 300 

S2 = 3999.9 = 4000 m 

 

Case – III:

 

Initial velocity = Vf = $13.333�{�}^{-1}$ 

Final velocity = Vi = 0 $�{�}^{-1}$ 

Time = t = 3 minutes = $3\times 60$ = 180 s

S3 = $���\times$ t 

S3 =$\frac{(��+��)}{2}\times$ 180 

S3 = $\mathrm{.}333+\frac{0}{2}\times$ 180

S3 = $6.6665\times 180$

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 $�{�}^{-1}$)

Difficulty: Medium

Solution:     

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

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

 Since,      ${�}_{�}={�}_{�}+�\times �$ 
                 ${�}_{�}={�}_{�}\text{–}�\times �$                                
                 ${�}_{�}=0\text{–}(-10)\times 3$
                 ${�}_{�}=30�{�}^{-1}$ 

Now using 3rd equation of motion

                  2aS = ${�}_{�}^2\text{–}{�}_{�}^2$ 
                  S = ${�}_{�}^2\text{–}\frac{{�}_{�}^2}{2�}$
                  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 $��{ℎ}^{-1}$ to 48 $��{ℎ}^{-1}$ 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 = ${�}_{�}=96��{ℎ}^{-1}$ = $96\times \frac{1000}{1000}$ = $\frac{96000}{96000}�{�}^{-1}$
Final velocity = ${�}_{�}=48��{ℎ}^{-1}$ = 48\times $\frac{1000}{3600}$ = $\frac{48000}{3600}�{�}^{-1}$
Distance = S = 800 m 
Further Distance = S1 =?

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

$2��={�}_{�}^2\text{–}{�}_{�}^2$
$2\times �\times 800$ = $(\frac{48000}{3600}{)}^2$ – $(\frac{96000}{3600}{)}^2$ 
1600a = $(\frac{48000}{3600}{)}^2$ – $((2\times \frac{48000}{3600}{)}^2$        $(96000=2\times 48000)$
1600a = $(\frac{48000}{3600}{)}^2((1{)}^2\text{–}(2{)}^2)$        (taking $(\frac{48000}{36000})$ as common)
1600a = $(\frac{48000}{3600}{)}^2(1\text{–}4)$
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:
${�}_{�}$ = 0              ,           S2 =?
2aS = ${�}_{�}^2\text{–}{�}_{�}^2$
$-2(\frac{48000}{36000}{)}^2$ \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 = ${�}_{�}=96��{ℎ}^{-1}$ = $96\times \frac{1000}{3600}$ =$\frac{96000}{3600}�{�}^{-1}$
Final velocity = ${�}_{�}=0�{�}^{-1}$ 
a =$(\frac{-48000}{3600}{)}^2$ \times $\frac{3}{1600}�{�}^{-2}$ 
time = t =?

Or   ${�}_{�}$ = ${�}_{�}$ + at 
Or   at = ${�}_{�}\text{–}{�}_{�}$

        t = ${�}_{�}\text{–}{�}_{\frac{{}_{�}}{�}}$
        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