Question

can anyone explain? ​

Answer 1

SOLUTION

Let 't1' be the time for which it accelerates, 'v' be the velocity after his tome. Now,when the acceleration is alpha.

$= >v = 0 + \alpha t1 \\ = > v = \alpha t1 \\ = > t1 = \frac{v}{ \alpha }$

Then it decelerates at rate Beta & stops at t2.

so,

$0 = v - \beta t2 \\ = > v = \beta t2 \\ = > t2 = \frac{v}{ \beta }$

Now,

$t = t1 + t2 = ( \frac{v}{ \alpha } ) + ( \frac{ v}{ \beta } ) = v( \frac{ \alpha + \beta }{ \alpha \beta } ) \\ \\ = > v = t( \frac{ \alpha \beta }{ \alpha + \beta } )$

Now,

Distance travelled in time t1 is found as,

${v}^{2} = {0}^{2} + 2 \alpha s1 \\ = > s1 = \frac{ {v}^{2} }{2 \alpha }$

Distance travelled in time t2 is found as,

${0}^{2} = {v}^{2} - 2 \beta s2 \\ = > s2 = \frac{ {v}^{2} }{2 \beta }$

Total distance travelled is:

S= s1+ s2

=) S= v^2 (2alpha)+ v^2 /(2beta)

=)S= [v^2/2 ] [(1/alpha+ 1/beta]

=)S= [v^2/2] [(alpha+beta)/alpha.beta]

=)S= 1/2[t[alpha.beta/alpha+beta]]^2 [(alpha + beta)/alpha.beta]

=)S= 1/2t[alpha.beta/(Alpha+beta)]

hope it helps ☺️⬆️