A truck running at 72 km/h is broght to rest over a distance of 25 m.calculate the retardation and the time for which the brakes are applied
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According to the question , Initial velocity of the truck = 72 km/h = 20 m/s Final velocity = 0 ( Body comes to rest ) Distance travelled = 25 m Retardation = ? Time taken = ? So ,Acceleration can be find out by the third equation of motion 2aS = v square - u square 2a X 25 = 0 square - 20 square 50a = 0 - 400 a = -400/5 Retardation = -8 m/s-2 Time taken can be find out by the first equation of motion, v = u + at = v-u /at = 0 - 20 / -8t = -20/-8t = 5/2t = Time taken = 2.5 seconds
Answered by
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Truck is running with speed = 72km/h
= 72 × 5/18 m/sec
= 20 m/s
Truck comes to rest after traveling 25 m .
hence, final velocity of truck = 0 m/s
use, kinematics formula,
V² = u² + 2as
here,
V = 0
u = 20 m/s
S = 25 m
0 = (20)² + 2 × a × 25
-(20)² = 50 × a
-400 = 50 × a
a = -8 m/s²
hence, retardation = - 8 m/s²
again,
use , kinematics formula,
V = u + at
here,
V = 0
u = 20 m/s
a = -8 m/s²
0 = 20 + (-8)t
t = 20/8 = 2.5sec
hence, time taken = 2.5 sec
= 72 × 5/18 m/sec
= 20 m/s
Truck comes to rest after traveling 25 m .
hence, final velocity of truck = 0 m/s
use, kinematics formula,
V² = u² + 2as
here,
V = 0
u = 20 m/s
S = 25 m
0 = (20)² + 2 × a × 25
-(20)² = 50 × a
-400 = 50 × a
a = -8 m/s²
hence, retardation = - 8 m/s²
again,
use , kinematics formula,
V = u + at
here,
V = 0
u = 20 m/s
a = -8 m/s²
0 = 20 + (-8)t
t = 20/8 = 2.5sec
hence, time taken = 2.5 sec
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