Question

17£®´òµã¼ÆʱÆ÷´òµãµÄʱ¼ä¼ä¸ôÊÇ0.02s£¬Ñ¡È¡Ö½´øÉϵÄ0¡¢1¡¢2¡¢3¡¢¡­Îª¼ÆÊýµã£¬ÏàÁÚÁ½¸ö¼ÆÊýµãÖ®¼ä»¹Ê¡ÂÔÁË4¸öµã£¬²âµÃ1¡¢2Á½µã¼äµÄ¾àÀëÊÇ7cm£¬3¡¢4Á½µã¼äµÄ¾àÀëÊÇ1.5 cm£¬ÔòÏàÁÚÁ½¸ö¼ÆÊýµãÖ®¼äµÄʱ¼ä¼ä¸ôΪt=0.1s£¬ÈôÖ½´øµÄÔ˶¯ÊÇÔȱäËÙÖ±ÏßÔ˶¯£¬2¡¢3Á½µã¼äµÄ¾àÀëÊÇ11cm£¬Õâ¶ÎÖ½´øµÄ¼ÓËٶȴóСÊÇ4m/s²£¬´òϵã2ʱֽ´øµÄ˲ʱËÙ¶ÈÊÇ0.9m/s£¬´òϵãOʱµÄËÙ¶ÈÊÇ0.1m/s£®

Answer 1

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½â´ð ½â£ºÏàÁÚÁ½¸ö¼ÆÊýµãʱ¼ä¼ä¸ôT=5¡Á0.02s=0.1s
ÈôÖ½´øµÄÔ˶¯ÊÇÔȱäËÙÖ±ÏßÔ˶¯£¬ÏàÁÚÏàµÈʱ¼äµÄλÒƲîÊǺãÁ¿${x}_{23}^{\;}-{x}_{12}^{\;}={x}_{34}^{\;}-{x}_{23}^{\;}$
´úÈ룺${x}_{23}^{\;}-7cm=15cm-{x}_{23}^{\;}$
½âµÃ£º${x}_{23}^{\;}=11cm$
ÏàÁÚÏàµÈʱ¼äλÒƲî$¡÷x={x}_{23}^{\;}-{x}_{12}^{\;}=11cm-7cm=4cm$=0.04m
¸ù¾Ý$¡÷x=a{T}_{\;}^{2}$
½âµÃ£º$a=\frac{¡÷x}{{T}_{\;}^{2}}=\frac{0.04m}{£¨0.1s£©_{\;}^{2}}=4m/{s}_{\;}^{2}$
´ò¼ÆÊýµã2µÄËÙ¶È${v}_{2}^{\;}=\frac{{x}_{13}^{\;}}{2T}=\frac{7cm+11cm}{2¡Á0.1s}=\frac{0.07m+0.11m}{0.2s}$=0.9m/s
${v}_{2}^{\;}={v}_{0}^{\;}+a•2T$
´úÈ룺$0.9={v}_{0}^{\;}+4¡Á0.2$
½âµÃ£º${v}_{0}^{\;}=0.1m/s$
¹Ê´ð°¸Îª£º0.1          11           4            0.9              0.1

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