Question

If brakes are applied to a car already
moving with a speed of 72 km/h to reduce it to the speed of 18 km/h in 10 seconds, calculate the retardation and the distance covered in the
10 seconds.​

Answer 1

$\huge{\underline{\underline{\sf{Answer :}}}}$

From the Question,

• Initial Velocity,u = 72Km/h

• Final Velocity,v = 18Km/h

• Time taken,t = 10 seconds

Here,

• All the quantities aren't in their SI units

Multiplying with 5/18,we get:

Initial Velocity,u = 20 m/s

Final Velocity,v = 5 m/s

Using the relation,

$\huge{ \sf{v = u + at}}$

Substituting the appropriate values,we get:

$\sf{5 = 20 + 10a}$

$\implies \ \sf{10a = -15}$

$\implies \ \sf{a = - \frac{15}{10}}$

$\implies \ \sf{a = -1.5 m{s}^{-2}}$

Using the relation,

$\huge{\sf{s = ut + \frac{1}{2}a{t}^{2}}}$

Putting the values,

s = (20)(10) + ½(-1.5)(10)²

»s = 200 - ½(15)(10)

»s = 200 - 75

»s = 125m

Car travels for a distance of 125m before coming to rest

Answer 2

$\huge{\underline{\underline{.........Answer.........}}}$

$\huge{\underline{Given:-}}$

v = 18 km/h

u = 72 km/h

t = 10 seconds

a = ?

S = ?

$\huge{\underline{Explanation:-}}$

Firstly converting km/h to m/s.

72 km/h = 72 × 5/18 m/s = 20 m/s.

18 km/h = 18 × 5/18 m/s = 5 m/s.

• Case-1

acceleration =?.

From first kinematics equation.

$\huge{\boxed{\boxed{a = \frac{v - u}{t}}}}$

Substituting the values.

${a = \frac{5 - 20}{10}}$

${a = \frac{- 15}{10}}$

${a = -1.5 m/s^2}$

$\huge{\boxed{\boxed{a = -1.5 m/s^2}}}$

So, deceleration of car is 1.5 m/s²

• Case-2

Distance = ?.

From second kinematic equation.

$\huge{\boxed{\boxed{S = ut + \frac{1}{2}a{t}^2}}}$

Substituting the values

${S = 20 \times 10 + \frac{1}{2} \times -1.5 \times 100}$

${S = 200 - 75}$

${S = 125 meters}$

$\huge{\boxed{\boxed{S = 125 meters}}}$

so,Distance covered in the period of retardation is 125 meters.