Question

a driver travelling at speed 36 km/hr sees the light turn red at the intersection. if his reaction time is 0.6s and then the car deaccelerate at 4m/s2. Find the stopping distance of the car
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Answer 1

Given

• Initial Velocity = 36 km/h
• Final Velocity = 0 m/s
• Time = 0.6 sec
• Retardation = 4 m/s²

To Find

• Distance Covered by the car

Solution

● First we shall convert the initial velocity from km/h to m/s and then use the second equation of motion to find the distance covered

✭ Initial Velocity :

→ Initial Velocity = 36 km/h

→ 36 × 5/18

→ 2 × 5

→ Initial Velocity = 10 m/s

━━━━━━━━━━━━━━━━━━━━

✭ Distance Covered

→ s = ut + ½at²

→ s = 10 × 0.6 + ½ × 4 × 0.6²

→ s = 6 + 2 × 0.36

→ s = 6 + 0.72

→ Distance = 6.72 m

• Note : Here we may also use the third equation of motion to find the distance covered and both the cases would give the same answer

Answer 2

Answer:

Given :-

• A driver travelling at a speed of 36 km/hr sees the light turn red at the intersection and the time is 0.6 s and the car deaccelerates at 4 m/s².

To Find :-

• What is the distance travelled by the car.

Formula Used :-

$\boxed{\bold{\large{s\: =\: ut\: +\: \dfrac{1}{2} a{t}^{2}}}}$

where,

• s = Distance
• u = Initial velocity
• t = Time
• a = Acceleration

Solution :-

Given :

◆ Initial velocity = 36 km/h = 36 km/s = $36 \times \dfrac{5}{18}$ m/s = 10 m/s

◆ Final velocity = 0 m/s

◆ Time = 0.6 seconds

◆ Acceleration = 4 m/s²

According to the question by using the formula we get,

⇒ s = $10 \times 0.6 + \dfrac{1}{2} \times 4 \times {(0.6)}^{2}$

⇒ s = $10 \times 0.6 + \dfrac{1}{2} \times 4 \times 0.36$

⇒ s = $10 \times 0.6 + 2 \times 0.36$

⇒ s = $6 + 0.72$

➠ s = 6.72 m

$\therefore$ The distance travelled by a car is 6.72 m .