9. In the given figure, Sis the mid-point of side QR of A PQR. PS is extended
to T such that PS = ST. Prove that PT bisects QR.
Answers
Step-by-step explanation:
Mass of the bus, ( m ) ⟿ 7000 kg
Time taken by the bus to stop, ( t ) ⟿ 4 second
Initial velocity of the bus, ( u ) ⟿ 36 km/ hr
ㅤㅤㅤㅤㅤㅤㅤㅤ⟿ \large \: \sf { { \large{\cancel{36}}}\times \large\frac{ 10 \cancel0 \cancel0 \: \red m \: \: }{ \cancel{36} \cancel0 \cancel0 \: \red s \: \: } }
36
×
36
0
0
s
10
0
0
m
ㅤㅤㅤㅤㅤㅤㅤㅤ⟿ 10 m/s
Final velocity of the bus ⟿ 0
ㅤㅤㅤㅤ( Because it comes to rest )
Acceleration of the bus ⟿ \large \sf\frac{F}{m}
m
F
\begin{gathered} \\ \\ \end{gathered}
\begin{gathered} \\ \huge\mathfrak {\underline \purple{ \: \: \: \: \: \: \: \: \: formula \mapsto \: \: \: \: \: \: \: \: }} \\ \\ \\ \\ \large \bigstar \sf \boxed{ \bf \red {v = u + at}} \bigstar \\ \end{gathered}
formula↦
★
v=u+at
★
\begin{gathered}\\ \huge\mathfrak {\underline \purple{ \: \: \: \: \: \: \: \: Solution \mapsto \: \: \: \: \: \: \: \: }} \\ \\ \end{gathered}
Solution↦
Putting the Values ➲
\begin{gathered} \large \sf \longmapsto 0 = 10 + (\frac{f}{m}) \times 4 \\ \\ \large \sf \longmapsto 0 = 10 + ( \frac{F}{7000}) \times 4 \\ \\ \large \sf \longmapsto 0 = 10 + \frac{4F}{7000} \\ \\ \large \sf \longmapsto 0 = \frac{70000 + 4F}{7000} \\ \\ \large \sf \longmapsto 0 = 70000 + 4F \\ \\ \large \sf \longmapsto - 70000 = 4 \: F \\ \\ \large \sf \longmapsto \: F = \frac{ - 70000}{4} \\ \\ \large \sf \longmapsto \boxed{ F= - 17500 \: \red N} \\ \\ \end{gathered}
⟼0=10+(
m
f
)×4
⟼0=10+(
7000
F
)×4
⟼0=10+
7000
4F
⟼0=
7000
70000+4F
⟼0=70000+4F
⟼−70000=4F
⟼F=
4
−70000
⟼
F=−17500N
Therefore , the force applied to stop the bus = -17500 N
Answer:
PS is extended to T such that PS = ST. Prove that PT bisects QR. .