Question

L & AIIMS
Motion in a Straight Line
5
04. A car travelling with a velocity of 80 km/h slowed down to 44 km/h in 15 s. The retardation is
(1) 0.67 m/s
(2) 1 m/s?
(4) 1.5 m/s2
(3) 1.25 m/s2
driving car with a speed 72 km/h, suddenly sees a boy crossing the road. If the distance mo​

Answer 1

» A car traveling with a velocity of 80 km/hr

Car is in motion means it's initial velocity (from starting) is 80 km/hr.

• u = 80 km/hr.

=> $\dfrac{80\:\times\:1000}{3600}$

=> 22.22 m/s.

_______________________________

» The car slowed down to 44 km/hr

Means the final velocity of the car is 44 km/hr in 15 sec.

• v = 44 km/hr.

=> $\dfrac{44\:\times\:1000}{3600}$

=> 12.22 m/s

• t (time) = 15 sec

______________________________

» We have to find the retardation means (a).

From first law of Equation.

v = u + at

We know the values of v, u, and t.

Put them in above equation.

=> 12.22 = 22.22 + a (15)

=> 12.22 - 22.22 = 15a

=> - 10.00 = 15a

=> a = $\dfrac{-10.00}{15}$

=> a = - 0.6667

=> a = - 0.67 m/s² (approx.)

_______________________________

The retardation is 0.67 m/s²

Option (1)

______________ [ ANSWER ]

Answer 2

Answer:

Retardedation = 0.67 m/s²

Step by step explanations :

given that,

A car travelling with a velocity of 80 km/h slowed down to 44 km/h in 15 s.

here,

initial velocity of the car = 80 km/h

= 80 × 1000 m/3600 s

= 200/9 m /s

it slowed down

and final velocity = 44 km/h

= 44 × 1000m/3600 s

= 110/9 m/s

given time taken for retardation = 15 s

now,

we have,

initial velocity(u) = 200/9 m/s

final velocity(v) = 110/9 m/s

time taken(t) = 15 s

acceleration = a

by the equation of motion,

v = u + at

putting the values,

110/9 = 200/9 + a(15)

15a = 110/9 - 200/9

15a = -90/9

15a = -10

a = -10/15

a = -0.67 m/s²

Negation of acceleration shows the retardation

so,

Retardedation = 0.67 m/s²