Question

A cycle running on a road with velocity 5 m/s is stopped by applying brakes at the rate of 10 m/s^2 calculate the displacement and the time taken

Answer 1

Answer :

Given -

• u = 5 m/s
• v = 0 m/s
• Acceleration = -10 m/s²

To Find -

• Displacement ?
• Time taken ?

Solution -

⇒ v = u + at

⇒ 0 = 5 + -10*t

⇒ -5 = -10t

⇒ t = -5/-10

⇒ t = 1/2

⇒ t = 0.5 s

Time taken = 0.5 s.

Now, s = ut + ½at²

⇒ s = 5*0.5 + ½*-10*0.5²

⇒ s = 2.5 + -5*0.5*0.5

⇒ s = 2.5 - 1.25

⇒ s = 1.25 m

So, Displacement = 1.25 m.

Answer 2

Correct question :

A cycle running on a road with velocity 5 m/s is stopped by applying brakes retarding at the rate of 10 m/s² calculate the displacement and the time taken.

Answer :

Given that a cycle is running on a road with a velocity of 5 m/s is stopped by applying brakes , retarding at the rate of 10 m/s².

We have to find displacement covered and time taken to stop.

From the data :

Initial velocity (u) = 5 m/s

Acceleration (a) = -10 m/s² [∵Acceleration = - Retardation]

Final velocity (v) = 0 [As it stops]

We will use 1st equation of motion to find time taken to stop :

→ v = u + at

→ 0 = 5 + (-10)t

→ -5 = - 10t

→ 5 = 10t

→ t = 5/10

→ t = 0.5 s

∴ Time taken to stop = 0.5 seconds.

___________________________

___________________________

Now we will use 2nd equation of motion to find displacement :

→ s = ut + ½ at²

→ s = 5 × 0.5 + ½ × (-10) × (0.5)²

→ s = 2.5 - 5 × 0.25

→ s = 2.5 - 1.25

→ s = 1.25 m

∴ Displacement covered by cycle = 1.25 m