A tiger running 100 m race, accelerates for one third time of the total time and then moves with uniform speed. Then find the total time taken by the tiger to run 100 m if the acceleration of the tiger is 8${m}/{{s}^{2}}$.
Answer
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Hint: Using the formula for acceleration, obtain ratio of velocity and time for the one third time of the total time. Find the distance covered by the tiger in one third of total time using the Kinematics equation. Then, find the distance in the remaining two thirds of total time using the basic formula relating distance and speed. Adding these 2 distances will give the total distance. Substitute the values in that equation and get an expression. Divide it by the ratio of velocity-time expression you obtained earlier. Thus, you can calculate the total time taken by the tiger.
Formula used:
$a=\dfrac {v}{t}$
$s= ut+ \dfrac {1}{2} $
Complete answer:
Given: Acceleration (a)= 8${m}/{{s}^{2}}$
Total distance (d)= 100m
Let the speed of tiger be v
Total time taken be T
The speed- time equation is given by,
$a=\dfrac {v}{t}$
Substituting values in above equation we get,
$8=\dfrac {v}{t}$ ...(1)
But, initially the tiger accelerates one third of the total time.
$\therefore t= \dfrac{T}{3}$
Substituting this in the equation. (1) we get,
$8=\dfrac {3v}{T}$
Rearranging the above expression we get,
$ \dfrac {v}{T}= \dfrac {8}{3}$
Using Kinematics equation, for the one third of total time,
$s= ut+ \dfrac {1}{2} a{t}^{2}$
But u=0,
$\therefore s= \dfrac {1}{2} a{t}^{2}$
We know, $s=a\times t$. Thus, substituting that in above equation we get,
$s= \dfrac {1}{2} v\times t$ ...(2)
Now substituting the values in the equation. (2) we get,
$s= \dfrac {1}{2} \times v\dfrac {T}{3}$...(3)
Now, for the two third of the total time the tiger is moving with constant speed,
$\therefore s= v \times t$
Substituting the values in above equation we get,
$s= v \times \dfrac {2T}{3}$ ...(4)
Adding equation. (3) and (4) we get the total distance
$\Rightarrow v \times \dfrac {2T}{3} + \dfrac {1}{2}\times v\times \dfrac {T}{3}= 100$
$\Rightarrow \dfrac {2vT}{3} + \dfrac {vT}{6}=100$
$\Rightarrow \dfrac {5vT}{6}=100$
$\Rightarrow vt= \dfrac {100 \times 6}{5}$
$\Rightarrow vt=120 $...(5)
But, we know $ \dfrac {v}{T}= \dfrac {8}{3}$ ...(6)
Now, dividing the equation. (5) by (6) we get,
${T}_{2} = \dfrac {120 \times 3}{8}$
$\therefore {T}_{2}= 45$
Taking the square root on both the sides,
$\therefore T= 6.71 s$
Hence, total time taken by the tiger to run 100m is 6.71 sec.
Note:
It is a lengthy numerical. There are a number of equations so don’t panic. Mark those equations properly. Solve the numerical part by part. Take care while writing the equations that you don't miss any term.Take care of the direction of acceleration.
Formula used:
$a=\dfrac {v}{t}$
$s= ut+ \dfrac {1}{2} $
Complete answer:
Given: Acceleration (a)= 8${m}/{{s}^{2}}$
Total distance (d)= 100m
Let the speed of tiger be v
Total time taken be T
The speed- time equation is given by,
$a=\dfrac {v}{t}$
Substituting values in above equation we get,
$8=\dfrac {v}{t}$ ...(1)
But, initially the tiger accelerates one third of the total time.
$\therefore t= \dfrac{T}{3}$
Substituting this in the equation. (1) we get,
$8=\dfrac {3v}{T}$
Rearranging the above expression we get,
$ \dfrac {v}{T}= \dfrac {8}{3}$
Using Kinematics equation, for the one third of total time,
$s= ut+ \dfrac {1}{2} a{t}^{2}$
But u=0,
$\therefore s= \dfrac {1}{2} a{t}^{2}$
We know, $s=a\times t$. Thus, substituting that in above equation we get,
$s= \dfrac {1}{2} v\times t$ ...(2)
Now substituting the values in the equation. (2) we get,
$s= \dfrac {1}{2} \times v\dfrac {T}{3}$...(3)
Now, for the two third of the total time the tiger is moving with constant speed,
$\therefore s= v \times t$
Substituting the values in above equation we get,
$s= v \times \dfrac {2T}{3}$ ...(4)
Adding equation. (3) and (4) we get the total distance
$\Rightarrow v \times \dfrac {2T}{3} + \dfrac {1}{2}\times v\times \dfrac {T}{3}= 100$
$\Rightarrow \dfrac {2vT}{3} + \dfrac {vT}{6}=100$
$\Rightarrow \dfrac {5vT}{6}=100$
$\Rightarrow vt= \dfrac {100 \times 6}{5}$
$\Rightarrow vt=120 $...(5)
But, we know $ \dfrac {v}{T}= \dfrac {8}{3}$ ...(6)
Now, dividing the equation. (5) by (6) we get,
${T}_{2} = \dfrac {120 \times 3}{8}$
$\therefore {T}_{2}= 45$
Taking the square root on both the sides,
$\therefore T= 6.71 s$
Hence, total time taken by the tiger to run 100m is 6.71 sec.
Note:
It is a lengthy numerical. There are a number of equations so don’t panic. Mark those equations properly. Solve the numerical part by part. Take care while writing the equations that you don't miss any term.Take care of the direction of acceleration.
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