A car accelerates from rest at a constant rate 4m/s for sometime after which it accelerates at constant rate 3m/s to come to rest . if total time is 14 s then average speed of car is

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Solution

Dear Student,

accn = 4m/s²., retardation =3 m/s^{2

$v = 4 t . . . . . 1 e q n f o r t h e a c c n 0 = v - 3 (14 - t) . . . . . . 2 e q n f o r t h e r e t a r d a t i o n . 0 = 4 t - 3 (14 - t) 3 (14 - t) = 4 t 42 - 3 t = 4 t 7 t = 42 t = 6 s e c . a t t = 6 s e c t h e r e t a r d a t i o n w i l l s t a r t . d i s \tan c e t r a v e l l e d i n 6 s e c a c c n . S = \frac{1}{2} \times 4 \times 36 = 72 m d i s \tan c e t r a v e l l e d i n r e t a r d a t i o n i n 8 s e c . S' = 24 * 8 - \frac{1}{2} \times 3 \times 64 = 192 - 96 = 96 m . t o t a l d i s \tan c e t r a v e l l e d = 96 + 72 = 168 m a v e r a g e s p e e d = \frac{168}{14} = 12 m / s R e g a r d s$}

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