Question

A body moving with a speed of 36 km per hour in abroad to rest in 10 seconds what is the negative acceleration and the distance travelled by the body before coming rest ?

Answer 1

Answer:

Distance traveled = 50 m

Retardation = 1 m/s²

Explanation:

As per the provided information in the given question, we have :

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 0 m/s (As it was brought to rest)
• Time taken (t) = 10 s

We are asked to calculate acceleration and distance covered before coming to rest.

Finding acceleration :

As we know that, Acceleration is the rate of change in velocity. Negitive acceleration is given by – m/s², It is known as retardation.

By using the first equation of motion,

⇒ v = u + at

• Final velocity is denoted by v
• Initial velocity is denoted by u
• Acceleration is denoted by a
• Time is denoted by t.

⇒ 0 = 10 + 10a

⇒ 0 - 10 = 10a

⇒ 10a = -10

⇒ a = -10/10

⇒ a = -1 m/s²

∴ Retardation is 1 m/s² .

Finding distance travelled :

By using the third equation of motion,

⇒ v² - u² = 2as

• Final velocity is denoted by v
• Initial velocity is denoted by u
• Distance is denoted by s
• Acceleration is denoted by a

⇒ (0)² - (10)² = 2(-1)s

⇒ 0 - 100 = -2s

⇒ -2s = -100

⇒ s = $\rm\cancel{\dfrac{-100}{-2}}$

⇒ s = 50

∴ Distance travelled by the body before coming to rest is 50 m.

Answer 2

Hola ⚘⚘

Answer:

• 50m

Explanation:

• Speed = 36km/hr
• Initial velocity = 10 m/s
• Final velocity = 0 m/s
• Time taken = 10 s
• distance travelled by the body before coming rest = ?

• v = u + at

>> 0 = 10 + 10a

>> 0 - 10 = 10a

>> 10a = -10

>> a = -10/10

>> a = -1m/s² (negative acceleration)

• v²-u²=2as

>> (0)²-(10)²= 2(-1)s

>> 0-100=-2s

>> -2s = -100

>> s = -100/-2

>> s = 50m

hence, distance travelled by the body before coming rest is 50m