Numerical Problems

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 

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 = $20 ms^{-1}$  

Solution:           

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

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

Solution:

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 = $10 \times \frac{3600}{1000}= 36 kmh^{-1}$

Solution: 

Case – I:      (6000 m)

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 

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 

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