گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!

A train starts from rest at a station $A$ and travels in a straight line to station $B$, where it comes to rest.

The train moves with constant acceleration $0.025{\text{ }}m{\text{ }}{s^{ - 2}}$ for the first $600{\text{ }}s$, with constant speed for the next $2600{\text{ }}s$, and finally with constant deceleration $0.0375{\text{ }}m{\text{ }}{s^{ - 2}}$.

a) Find the total time taken for the train to travel from $A$ to $B$.

b) Sketch the velocity-time graph for the journey and find the distance $AB$.

c) The speed of the train $t$ seconds after leaving $A$ is $7.5{\text{ }}m{\text{ }}{s^{ - 1}}$. State the possible values of $t$.

پاسخ تشریحی :
نمایش پاسخ

a) $v\left( {600} \right) = 0.025 \times 600$

$0 = 15 - 0.0375{t_3}$

Total time is $3600{\text{ }}s$

b) For correct graph

[$d = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}\left( {2600 + 3600} \right) \times 15$ or

$d = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}0.025 \times {600^2} + 2600 \times 15 + $

${\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}0.0375 \times {400^2}$]

Distance is $46500$

c) Values of t are 300 and 3400

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!