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This is second of our many quizzes on Acceleration. Click on this to access more such competitive quizzes on Mechanics.

Acceleration Quiz 2

physics

Using the following figure which depicts the kinematics of a lizard while it's hunting for prey, determine the acceleration of the lizard between intervals A and B, B and C, and D and E $(\text{ all in}\, ms^{-2}).$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/9ee54f4e-60fa-4ded-8a60-ed2bb2f261127266312555983767875.png"/>

 Obtain the initial and final velocity in each interval and calculate the time period following which you can find the lizard’s acceleration in each case. 

  $\text{ Between A and B:}\\ u = 0\,ms^{-1}, v = 5\,ms^{-1}\,\text{and}\, \Delta t = 2-0=0\\ \text{ Acceleration of the lizard between A and B will be:}\\ a = \dfrac{v-u}{\Delta t}\\ \Rightarrow a = \dfrac{5-0}{2}\\ \therefore a= 2.5\,ms^{-2}\\ \text{ Between B and C:}\\ u = 5\,ms^{-1}, v = 5\,ms^{-1}\, \text{and}\, \Delta t = 5-2=3\\ \text{ Acceleration of the lizard between B and C will be:}\\ a = \dfrac{v-u}{\Delta t}\\ \Rightarrow a = \dfrac{5-5}{3}\\ \therefore a = 0\,ms^{-2}\\ \text{ Between D and E:}\\ u = 7\,ms^{-1},v = 0\,ms^{-1}\,\text{and}\, \Delta t = 9-6=3\\ \text{ Acceleration of the lizard between D and E will be:}\\ a = \dfrac{v-u}{\Delta t}\\ \Rightarrow a = \dfrac{0-7}{3}\\  \therefore a = -2.33\,ms^{-2}$

 Consider a pullback toy truck that gets propelled forward upon pulling it back a certain distance as shown below. Upon its release from point A, the truck moves at the rate of $2\,ms^{-1}$ at point O before coming to a complete stop at point B in $20\,s$. What is the distance OB through which the pullback truck gets propelled? If the same truck is pushed with a force of $2\,N$ at point O, what will be the net acceleration of the toy truck? Take mass of the truck as $m=0.5\,kg$<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fa1cbf17-fbb6-4e9a-b9a1-3dde3a71a15d5305284419954981714.png"/> 

First find the acceleration of the truck between O and B to find distance OB. Then find the acceleration on the truck due to the applied force and sum it with its inherent acceleration. 

  $\text{ Given that between O and B,} u = 2\,ms^{-1}, v = 0\,ms^{-1}\,\text{and}\,t=20\,s.\\ \text{ The acceleration of the truck will be,}\\ a =\dfrac{v-u}{t}\\ \Rightarrow a = \dfrac{0-2}{20}\\ \Rightarrow a = -0.1\,ms^{-2}\\ \text{ Distance OB is,}\\ OB= ut+\dfrac{1}{2}at^2\\ \Rightarrow OB= 2 \times 20 - \dfrac{1}{2} \times 0.1 \times 20^2\\ \Rightarrow OB = 40 – 20 = 20\,m\\ \text{ Now, the truck is given an additional push of F = 2N at O.}\\ \text{ The acceleration imparted to the truck of mass m = 0.5 kg}\\ \text{ by virtue of application of this force can be given as:}\\ a^{\prime}=\dfrac{F}{m}\\ \Rightarrow a^{\prime} = \dfrac{2}{0.5}\\ \Rightarrow  a^{\prime} = 4ms^{-2}\\ \text{ The net acceleration of the toy truck will be:}\\ a_{net} = a+a^{\prime}\\ \Rightarrow a_{net} = -0.1 + 4 \\ \therefore a_{net} = 3.9\,ms^{-2}$

 You are on a jungle safari through the African Savannah when a cheetah suddenly leaps in front of your truck. If the truck was initially cruising at a velocity of $15\,kmh^{-1}$ and took $3\,s$ to come to a complete stop, what was the acceleration of the truck? Now, as the driver slowly pulls the truck away from the cheetah, it begins to chase the truck. If the cheetah can accelerate at the rate of $5.71\,ms^{-2}$ in $3.5\,s$ from rest, what must be the minimum velocity of the truck in order to outrun the cheetah? 

The velocity that the truck must possess in order to outrun the cheetah will be the maximum final velocity that the cheetah can attain. Be cautious of the units used.  

 $ \text{ Given that initially for the truck,}\\ u = 15\,kmh^{-1} = 15 \times \dfrac{5}{18} = 4.17\,ms^{-1}\\ v=0\,ms^{-1}\,\text{and}\, t = 3\,s\\ \text{ The acceleration or retardation of the truck would be:}\\ a=\dfrac{v-u}{t}\\ \Rightarrow a = \dfrac{0-4.17}{3}\\  \Rightarrow a =  -1.39 \approx -1.4\,ms^{-2}\\ \text{ Now, for the cheetah, we have}\\ u=0\,ms^{-1}, a=5.71\,ms^{-2}\,\text{and}\, t=3.5\,s\\ \text{The minimum velocity that the truck must speed away}\\ \text{will be the final velocity that the cheetah can attain:}\\ v = u+a \times t\\ \Rightarrow v = 0+5.71 \times 3.5\\ \Rightarrow v = 19.985\,ms^{-1}\\ \Rightarrow v = 19.985 \times \dfrac{18}{5}\\ \therefore v = 71.95 \approx 72\,kmh^{-1}$

Falcon 9 is SpaceX’s partially reusable two-stage rocket capable of transporting people and payload alike. Its reusability arises from the fact that the first stage of the rocket separates from the second stage in space and re-enters the atmosphere and lands vertically on a drone ship on the Earth. If the Falcon 9 weighs $5 \times 10^6\,kg$ and the thrust acting on it is $6000 \times 10^2\,kN$, what is the net acceleration of the rocket? $(\text{Take g} = 10\,ms^{-2})$ 

 Find the net force acting on the rocket due to its weight and thrust and then arrive at the acceleration of the rocket.  

$ \text{ Given that the mass of the rocket of m} = 5 \times 10^5\,kg.\\ \text{The weight of the rocket acting downwards will be:}\\ W = mg\\ \Rightarrow W = 5 \times 10^6 \times 10\\ \Rightarrow W = 0.5 \times 10^8\,N\\ \text{ The thrust on the rocket has a magnitude of}\, T = 6 \times 10^8\,N\,\text{and acts upwards.}\\ \text{ Therefore, the net force acting on the rocket will be:}\\ F_{net} = T-W\\ \Rightarrow  F_{net} = 6 \times 10^8 – 0.5 \times 10^8\\ \Rightarrow  F_{net} = 5.5\times 10^{8}\,N\\ \text{ Therefore, the net acceleration of the rocket will be:}\\ a = \dfrac{F_{net}}{m}\\ \Rightarrow a = \dfrac{5.5\times 10^8}{5 \times 10^6}\\ \Rightarrow a = 110\,ms^{-2}\\ \Rightarrow a = 110 \times 10^{-3}\\ \therefore a= 0.11\,kms^{-2}$

 Imagine you’re in the Broadway Theatre watching a troop perform the Black Swan. One of the ballerinas takes the centre stage and begins to perform a Pirouette, where she spins on one foot by fully rotating her body. If she starts by executing one pirouette every two seconds and ends at 3 pirouettes every two seconds in a period of ten seconds, what would be her angular acceleration of revolution ? 

Begin by finding the ballerina’s initial and final velocity following which you can arrive at her angular acceleration in the given time. 

 $ \text{ Given that the ballerina initially executes 1 pirouette(revolution) in two seconds,}\\ \text{ the initial angular velocity of the ballerina will be:}\\ \omega_i = \dfrac{1 \times 2\pi}{2}\\ \Rightarrow\omega_i  = \pi\,rads^{-1}\\  \text{ She finally executes 3 pirouettes(revolutions) in two seconds,}\\ \text{ so her final angular velocity will be:}\\ \omega_f = \dfrac{3 \times 2\pi}{2}\\ \Rightarrow\omega_f = 3\pi\,rads^{-1}\\ \text{ Therefore, her angular acceleration in a period of t=10 s can be given as:}\\ \alpha = \dfrac{\omega_2- \omega_1}{t}\\ \Rightarrow\alpha = \dfrac{3\pi - \pi}{10}\\ \therefore\alpha = 0.2\pi\,rads^{-2}$