A car moving along a straight road with a
speed of 120 km/hr, is brought to rest by
applying brakes. The car covers a distance
of 100 m before it stops. Calculate (i) the
average retardation of the car (ii) time
taken by the car to come to rest.
[Ans: 50/9 m/sec, 6 sec]
Answers
Answer:
Answer:
(i)a=-72000 km/hr^2a=−72000km/hr
2
(ii) 6 sec
Explanation:
We are given that a car moving along a straight road with a speed 120 km/hr.
Final velocity=0 km/hr
Initial velocity=120 km/hr
Distance covered by the car=100 m=\frac{100}{1000}=0.1 Km
1000
100
=0.1Km
We have to (i) the average retardation of the car
(ii) time taken by the car to come to rest
We know that
v^2-u^2=2asv
2
−u
2
=2as
Substitute the value then we get
0-(120)^2=2\times\frac{1}{10}a0−(120)
2
=2×
10
1
a
a=-14400\times 5 km/hr^2a=−14400×5km/hr
2
a=-72000 km/hr^2a=−72000km/hr
2
Retardation is an acceleration in opposite direction.
-72000 =\frac{v-u}{t}−72000=
t
v−u
substitute the value then we get
72000=\frac{-120}{t}72000=
t
−120
t=\frac{120}{72000}t=
72000
120
t=0.00167 hrt=0.00167hr
t=0.00167\times 3600=6 sect=0.00167×3600=6sec
Because 1 hr=3600 sec
Hence, the taken by the car to come to rest =6 sec.
Answer:
Explanation: v^2 - u^2 =2as .......(1)
s=100m u = 120 km/h v= 0 ........(car is at rest )
u=33.3m/s ........(approx)
putting these values we get
0- 1108.89= 2 a 100
a=-(11.08/2) ............(approx)
a=5.54
now v= u +at ..........(2)
t= -(u/a)
t =6.06 s .....(apprix)