Tiêu đề 3.-Phy-1st-Paper-Gotibidda-Solve Sheet.pdf ✅

4  Physics 1st Paper Chapter-3 14| x = 3t3 + 4t2 + 3t Øviv GKwU e ̄‘i miY wgUv‡i wb‡`©wkZ nq| 4 sec c‡i e ̄‘wUi Z¡iY wbY©q Ki| [KUET 07-08] mgvavb: x = 3t3 + 4t2 + 3t  v = dx dt = 9t2 + 8t + 3  a = dv dt = 18t + 8  4 sec ci Z¡iY = 18  4 + 8 = 80 ms–2 (Ans.) 15| GKwU e ̄‘KYvi †eM‡K vs = 0.10 ms–1 + (0.02 ms–1 )t 2 Øviv cÖKvk Kiv hvq| 2 sec I 5 sec Gi e ̄‘KYvi Mo Z¡iY KZ? [KUET 09-10] mgvavb: vs = 0.10ms–1 + (0.02ms–1 )t2 as = dvs dt = d dt [0.1 + 0.02 t2 ] as = 0.02  2t = 0.04t Mo Z¡iY, – a = a2 + a5 2 = (0.04  2) + (0.04  5) 2 = 0.08 + 0.2 2 = 0.14 ms–2 (Ans.) Alternative:  Mo Z¡iY, – a = v5 – v2 5 – 2 = 0.14 ms–2 (Ans.) GLv‡b, vt = 0.1 + 0.02t2 v2 = 0.18 ms–1 v5 = 0.6 ms–1 16| GKwU e ̄‘i Z¡iY ‘a’ m/sec2 mgq ‘t’ sec Gi mv‡_ a = 3t – 1 mgxKiY Abyhvqx cwiewZ©Z nq| t = 2 sec mg‡q e ̄‘wUi MwZ KZ n‡e? [RUET 11-12, BUET 07-08] mgvavb: a = dv dt  dv = adt   dv =  adt  v = 3  2 0 t dt –  2 0 dt = 3    t 2 2 2 0 – [t] 2 0 = 3 2  4 – 2 = 4 ms–1 (Ans.) 17| wgUv‡i cÖKvwkZ GKwU e ̄‘i Ae ̄’vb x(t) = 16t – 3t3 †hLv‡b mgq t †m‡K‡Û cÖKvwkZ| e ̄‘wU ÿwY‡Ki Rb ̈ w ̄’ive ̄’vq _v‡K hLb t Gi gvbÑ [BUET 12-13] mgvavb: x(t) = 16t – 3t3  v = dx dt = 16 – 9t2 w ̄’i _vK‡j, v = 0  16 – 9t2 = 0  t = 16 9 = 1.3333 s  1.30 s (Ans.) 18| †Kv‡bv e ̄‘i Ae ̄’v‡bi mgxKiY x = 3t2 – t 3 n‡j e ̄‘wU abvZ¥K x Aÿ eivei m‡e©v”P KZ `~i hv‡e? mgvavb: v = dx dt = 6t – 3t2 v Gi gvb k~b ̈ nIqvi AvM ch©šÍ e ̄‘wU x A‡ÿi abvZ¥K w`‡K hv‡e|  6t – 3t2 = 0  t = 2  m‡e©v”P miY, x = 3  2 2 – 2 3 = 4m (Ans.) 19| cÖ`Ë †jLwP‡Îi a – t MÖvd †_‡K v – t I x – t MÖvd AuvK| 0 1 0 5 sec Z¡iY (ms–2 ) 10 mgvavb: MwZkxj e ̄‘wUiwUi Z¡iY Amg hvi mgxKiYÑ a(t) = 2t; (0  t  5) Ges a(t) = 10; (5 < t < 10) GLb, (v – t) †jL Gi Rb ̈Ñ (0  t  5) sec mgq Gi g‡a ̈ e ̄‘wUi †eM, v =  5 0 adt =  5 0 2tdt =     2t2 2 5 0 = 25 ms–1 Abyiƒc fv‡e, (5 < t < 10) sec Gi g‡a ̈ t = 10 sec G e ̄‘wUi †eM, v = u + at = 25 + (10  5) = 75 ms–1  v – t †jLwPÎÑ 0 10 t(s) v (m/s) 5 25 75 x – t †jL Gi Rb ̈: (0 < t < 5) sec G e ̄‘wU KZ...©K AwZμvšÍ c_, x(5) =  5 0 vdt –  5 0 2tdt =     t 3 3 5 0 = 41.67 m Abyiƒcfv‡e, (5 < t < 10) sec G e ̄‘wUi AwZμvšÍ `~iZ¡ G mg‡qi g‡a ̈ v – t †jL G Ave× †ÿÎd‡ji mgvb| ZvB (5 < t < 10) sec G e ̄‘wUi AwZμvšÍ `~iZ¡, x = 1 2  (25 + 75)  5 = 250 m.  x – t †jLwPÎÑ 0 10 t(s) x (m) 5 41.67 291.67