Question

A CAR ACCELERATES FROM REST AT 5M/S2 AND THEN RETARDS TO REST AT 3M/S2.THE MAXIMUM VELOCITY OF THE CAR IS 30M/S.WHAT IS THE DISTANCE COVERED BY THE CAR DURING ENTIRE JOURNEY

Answer 1

$\textbf{ Hello Mate! }$

Here, we will go for 2 cases.

Case 1 is case when car acclerated 5 m/s^2 from rest ( u = 0 ) and attained 30 m/s velocity ( v ). Let's keep values in equation.

${v}^{2} - {u}^{2} = 2as \\ {30}^{2} - {0}^{2} = 2(5)(s) \\ 900 = 10s \\ 90 \: m = s$

Case 2 is case when car was moving with velocity of 30 m/s ( u ) but suddely applied breaks for retardation of 3 m/s^2 to come to rest ( v = 0 ). Let's keep values.

${v}^{2} - {u}^{2} = 2as \\ {0}^{2} - {30}^{2} = 2( - 3)(s) \\ - 900 = - 6s \\ 150 \: m= s$

Total sistance covered = Case 1 + Case 2

= 90 m + 150 m

= 240 m

$\boxed{ \textsf{ \red{ Hence,\:distance\:covered\:is\:240\:m }}}$

Have great future ahead!