Question

Question 3.6 A car moving along a straight highway with a speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?

Class XI Physics Motion In A Straight Line Page 56

Answer 1

Initial velocity of the car, u = 126 km/h = 35 m/s
Final velocity of the car, v = 0
Distance covered by the car before coming to rest, s = 200 m
Retardation produced in the car = a
From third equation of motion, a can be calculated as:
v2 – u2 = 2as
(0)2 – (35)2 = 2 × a × 200
a = – 35 × 35 / 2 × 200 = – 3.06 ms-2
From first equation of motion, time (t) taken by the car to stop can be obtained as:
v = u + at
t = (v – u) / a = (- 35) / (-3.06) = 11.44 s

Answer 2

Initial Speed of the car =  126 km/h × 5/18 =  35 m/sec ( changed in m per sec because distance is given in metre)  
Final velocity of the car = 0 m/sec
Distance covered while retarding =  200 m
Let the acceleration be a.
 
from the third equation of motion, we know that
     v² = u² + 2as
⇒  0² = 35² + 2×a× 200
⇒ -1225/400 = a
 ∴ a =  -3.06 m/sec²

so, the retardation of the car is 3.06 m/sec².
the negative sign symbolise that it is negative acceleration or retardation.

now, for finding the time.
we know from the first equation of motion that :
  v = at + u
⇒  -35/-3.06 = t
∴ t =  11.44 sec (approximately)

so, time taken by the car to stop is 11.44 secs.