Question

✔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.

PROPER ANSWER NEEDED...

give full explanation..

​

Answer 1

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).

I hope u stand it clearly. ..plzz mark me at brainlist

Answer 2

$\huge\sf{Solution:}$

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:

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

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

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

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