Physics, asked by Anonymous, 11 months ago

✔100 POINTS✔

a car accelerates from rest at constant rate (alpha) for some time, after which it decelerate at a constant rate (beta), to come to rest. if the volume elpased at t. evalute (1) max. velocity attained (2) total distance travelled.

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Answers

Answered by aliya346
8

Answer:

for the first leg of journey

v=0+at1

so. v=at1

for the second leg of journey of duration t-t1

0=v-beta (t-t1)

0= at1- beta T +beta T

t1=betaT/(a+b)

v= at1= a beta T/( a+b).

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Answered by Anonymous
30

\huge\sf{Solution:}

Let the car accelerate for time t_{1} travelling distance S_{1} and acquiring maximum velocity v.

Note: Check attachment no. 1

After this car decelerates for time t_{1} to come to rest.

Note: Check attachment no. 2

Also:

Total distance travelled:

\implies \sf{S = S_{1} + S_{2}}

\implies \sf{S  =  \frac{ {v}^{2} }{2} ( \frac{1}{ \alpha }  +  \frac{1}{ \beta } )}

\implies \sf{S =  \frac{1}{2} (  \frac{ \alpha  \beta t}{ \alpha  +  \beta } ) ^{2} ( \frac{ \alpha  +  \beta }{ \alpha  \beta } )}

\implies \boxed{\sf{S =  \frac{1}{2} ( \frac{ \alpha  \beta }{ \alpha  +  \beta } ) {t}^{2}}}

Attachments:

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