Question

A scooter moving at a speed of 10m/s is stopped by applying breakes which produce a uniform acceleration of, -0.5 m/s². How much distance will be covered by the scooter before it stops.​

Answer 1

Explanation:

Solution -

Solution -We have,

Solution -We have,Initial speed (u) = 10m/s

Solution -We have,Initial speed (u) = 10m/sFinal speed (v) = 0

Solution -We have,Initial speed (u) = 10m/sFinal speed (v) = 0Acceleration (a) = -0.5m/s²

Solution -We have,Initial speed (u) = 10m/sFinal speed (v) = 0Acceleration (a) = -0.5m/s²⠀

= u^2 + 2as}}}}}⟼

v² =u² +2as

+2as

+2as

+2as ⠀

+2as ⠀Putting values

{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}⇢(0) 2

{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}⇢(0) 2 =(10)

{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}⇢(0) 2 =(10) 2

{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}⇢(0) 2 =(10) 2 +2×(−0.5)×s

{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}⇢(0) 2 =(10) 2 +2×(−0.5)×s⠀

{0 = 100 - 1 \times s}⇢0=100−1×s

{0 = 100 - s}⇢0=100−s

{0 = 100 - s}⇢0=100−s⠀

{0 = 100 - s}⇢0=100−s⠀{s = 100\ m}⇢s=100m

m}⇢s=100m⠀

distance covered is 100 metres

metresThus,thedistancecoveredis100metres.

metresThus,thedistancecoveredis100metres.

Answer 2

Solution -

We have,

• Initial speed (u) = 10m/s
• Final speed (v) = 0
• Acceleration (a) = -0.5m/s²

⠀

We have to calculate the distance covered (s) by the scooter before it stops.

⠀

Using third equation of motion

$\large{\bf{\longmapsto{\boxed{\pink{v^2 = u^2 + 2as}}}}}$

⠀

Putting values

$\tt\dashrightarrow{(0)^2 = (10)^2 + 2 \times (-0.5) \times s}$

⠀

$\tt\dashrightarrow{0 = 100 - 1 \times s}$

⠀

$\tt\dashrightarrow{0 = 100 - s}$

⠀

$\bf\dashrightarrow{s = 100\: m}$

⠀

$\: \: \: \: \: \underline{\sf{Thus,\: the\: distance\: covered\: is\: 100\: metres.}}$