Question

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

Calculate :
8.1. The velocity after 12 seconds.
8.2. The displacement when the velocity is 18ms?​

Answer 1

• 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

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

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

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

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

$\\ \sf\longmapsto s=29.4m$

Answer 2

Answer:

Question 8.1

• from first newtons equation of motion:

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

• u → 0

• a → 5.5 m/s²

• t → 12 s

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

Question 8.2

• 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

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