Question

A car moving along a straight highway with speed of 226 km/hr is broughtto astop within a distance of 800 m . Wht is the retardation of the car (assumed uniform) and how long does it take for the car to stop.​

Answer 1

GIVEN :-

• Initial velocity (u) = 226km/h
• Final velocity (v) = 0m/s
• Distance (s) = 800m
• Retardation (negative acceleration) is uniform.

TO FIND :-

• Retardation (-a).

SOLUTION :-

Initial velocity (u) = 226km/h

We will convert it into m/s.

[ For conversion of km/h to m/s , we hav to multiply by (5/18) ]

Initial velocity in m/s will be [226 × (5/18) ]

Initial velocity (u) = 62.77m/s

By 3rd Kinematical equation ,

★ v² = u² + 2as

Putting values , we get...

→ 0² = (62.77)² + 2(a)(800)

→ 0 = 3940.5 + 1600a

→ 1600a = -3940.5

→ a = -3940.5/1600

→ a = -2.45 m/s²

→ -a = 2.45m/s²

Hence , retardation of the car is 2.45m/s².

MORE TO KNOW :-

♦ 1st Kinematical equation :-

★ v = u + at

♦ 2nd Kinematical equation :-

★ s = ut + (1/2)at²

Here ,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance
• t = Time taken

♦ These Kinematical equations are only applicable when acceleration is uniform.