
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
574.8k+ views
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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

