Consider a car initially at rest, starts to move along a straight road first with acceleration `5 m//s^2`, then with uniform velocity and finally, decelerating at `5 m //s^2` , before coming to a stop . Total time taken from starts to end is t= 25 s . If the average velocity during that time is 72 km/ hr , the car moved with uniform velocity for a time of

Options

A. 15 s
B. 30 s
C. 155 s
D. 2 s

Answer

A

Explanation

Given Acceleration of car, `a= 5m//s^2`
deceleration of car `a= 5m//s^2`, total time taken from start of end is ,t= 25 s and average velocity of car,
`V_("avg") = 72 km//hr =20 m//s (because 1 (Km)/(hr) = 5/(18) m//s)`
Since , `V_("avg") = ("total displacement")/("total time taken") `
2t , total time taken by the car during acceleration and deceleration
`V_("avg") =20 =(d_t+d_(25-2t)+d_t)/(25)`
= `(2d_t+d_(25-2t))/(25)`
since , `d_t =0+1/2 at^2=1/2at^2=5/2t^2 and d_(25-2t)=V_("uni") (25-2t)`
where , `v_("uni") =5t`
now, `V_("avg") =20 (2(5/2t^2)+5t(25-2t))/(25)`
`therefore 20 xx 25 =5t^2 +5t (25-2t)`
`rArr 500 = 5t^2+125t -10 t^2`
`rArr t^2-25t+100 =0`
So, it given t=20 and 5 s.
Hence, the time of uniform motion,
`t_(20) =25 -2t=25-2xx20=-15s`
(`because` Not possible)
or ` t_5 =25 -10 =15s`
So, the correct option is (a)