A lift ascends from rest with a uniform acceleration of 4m/m², then it moves with uniform velocity and finally comes to rest with a uniform retardation of 4 m/s².If the total distance covered during ascending of the lift is 28m and the total time taken for ascending the lift is 8s respectively, then find the time in which the lift moves with uniform velocity.Also find its uniform velocity

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Dear Student ,
$S i n c e t h e a c c e l e r a t i o n a n d t h e d e c e l e r a t i o n a r e t h e s a m e, t h e t i m e s p e n t a c c e l e r a t i n g a n d d e c e l e r a t i n g a r e t h e s a m e . S o, s = 2 \times \frac{1}{2} a t^{2} + v (8 - 2 t) F u r t h e r m o r e, v = a t, s o s = a t^{2} + a t (8 - 2 t) = a t^{2} + 8 a t - 2 a t^{2} = 8 a t - a t^{2} W e' r e g i v e n s = 28 a n d a = 4 S o 28 = 8 \times 4 \times t - 4 \times t^{2} = 32 t - 4 t^{2} \to d i v i d e b y 4 a n d r e a r r a n g e t^{2} - 32 t + 28 = 0 q u a d r a t i c w i t h r o o t s a t t = 31 s (t o o l a r g e) a n d t = 0.90 s \to t i m e t o r e a c h u n i f o r m v e l o c i t y v = 4 m / s^{2} \times 0.90 s = 3.6 m / s \to u n i f o r m v e l o c i t y S o, s = 2 \times \frac{1}{2} \times 4 \times (0.9)^{2} + 3 \cdot 6 \times 6 \cdot 8 \approx 28 m$
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