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grigory [225]
3 years ago
11

8. A car starts from rest and accelerates at 5.5ms in an easterly direction.

Physics
2 answers:
snow_lady [41]3 years ago
8 0
  • Initial velocity=0m/s=u
  • Acceleration=a=5.5m/s^2

#8.1

  • t=12s

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+5.5(12)

\\ \sf\longmapsto v=66m/s

#8.2

  • u=0m/s
  • v=18m/s
  • a=5.5m/s^2

Use third equation of kinematics

\\ \sf\longmapsto v^2-u^2=2as

\\ \sf\longmapsto s=\dfrac{v^2-u^2}{2a}

\\ \sf\longmapsto s=\dfrac{18^2-0^2}{2(5.5)}

\\ \sf\longmapsto s=\dfrac{324}{11}

\\ \sf\longmapsto s=29.4m

wel3 years ago
4 0

Answer:

<u>Question</u><u> </u><u>8</u><u>.</u><u>1</u>

• from first newtons equation of motion:

\dashrightarrow \: { \tt{v = u + at}}

• u → 0

• a → 5.5 m/s²

• t → 12 s

\dashrightarrow \:  { \tt{v = 0 + (5.5 \times 12)}} \\  \\ { \boxed{ \tt{v = 66 \:  {ms}^{ - 1} }}}

<u>Question</u><u> </u><u>8</u><u>.</u><u>2</u>

• from third newton equation of motion:

{ \tt{ {v}^{2} =  {u}^{2}   + 2as }}

• s → displacement

• u → 0

• a → 5.5 m/s²

• v → 18 m/s

\dashrightarrow \: { \tt{ {18}^{2} =  {0}^{2}  + (2 \times 5.5 \times s) }} \\  \\ { \tt{324 = 11s}} \\  \\ { \boxed{ \tt{displacement = 29.5 \: m}}}

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