Question

20. A car accelerates from rest at a constant α for same time after which it decelerates at a constant rate β and comes to rest . If the total time elapsed is t, then the maximum acquired by the car is ​

Answer 1

Answer:

A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t, the maximum velocity acquired by the car is given by.

Explanation:

Answer :

A::B::C::D

Solution :

(a) Let the car accelerates for time t1 and decelerates for time t2. Then

t=t1+t2….(i)

and corresponding velocity-time graph will be as shown in fig.

(DCPV01CH6S01024_S01.png width=80%>Fromthegraph,alpha=slopeofl∈eOA=v_(max)/t_1ort_1=v_(max)/alpha....(ii)andbeta=−slopeofl∈eAB=v_(max)/t_2ort_2=v_(max)/beta...(iii)FromEqs.(i),(ii)and(iii),we≥tv_(max)/alpha+v_(max)/beta=torv_(max)((alpha+beta)/(alpha beta))=torv_(max)=(alpha beta t)/(alpha+beta)(b)Totaldistance=Totaldisplacement=areaunderv−tgraph=1/2xxtxxv_(max)=1/2xxtxx(alpha beta t)/(alpha+beta)orDistance= 1/2((alpha beta t)/(alpha+beta))`.

Answer 2

Answer:

HEY MATE HERE IS YOUR ANSWER....

Let maximum velocity=v 

Now,v=0+αt1

Similarly, 0=v−βt2

From the above equations we get,

t1=αv

t2=βvt1+t2=t=αv+βv

⇒v=α+βαβt

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