Q18. A car moving with s speed of 40 m/s can be stopped by applying brakes in 2 m. If the same acceleration exists, after how much distance in which a car with velocity 80 m/s can be stopped?
Answers
Answer:
when the velocity of car will 80 m/s
➪ distance travelled = 8 m
Step by step explanations :
given that,
A car moving with s speed of 40 m/s
can be stopped by
applying brakes in 2 m.
here,
initial velocity of the car ⛟ = 40 m/s
final velocity of the car = 0 m/s
[it comes to rest]
distance travelled after applying brakes
= 2 m
let the retardation produced by the brakes be a
now we have,
initial velocity(u) = 40 m/s
final velocity(v) = 0 m/s
distance travelled(s) = 2 m
acceleration(a) = a
by the equation of motion,
v² = u² + 2as
putting the values,
(0)² = 40² + 2a(2)
4a + 1600 = 0
4a = -1600
a = -1600/4
a = -400
Negation of acceleration shows the retardation
now,
given initial velocity of the car = 80 m/acceleration is the same
= -400 m/s
final velocity(v) = 0 m/s
again
v² = u² + 2as
putting the values,
(0)² = 80² + 2(-400)s
-800s + 6400 = 0
-800s = -6400
S = -6400/-800
S = 8 m
so,
when the velocity of car will 80 m/s
distance travelled = 8 m
Answer:
distance travelled = 8 m
Step by step explanations :
given,
for a car
initial velocity(u) = 40 m/s
final velocity(v) = 0 m/s
[brake applied to rest]
distance travelled(s) = 2 m
v² = u² + 2as
a = acceleration
0² = 40² + 2a(2)
4a = -1600
a = -1600/4
a = -400
initial velocity = 80 m/s
acceleration = -400 m/s
final velocity = 0 m/s
v² = u² + 2as
0² = 80² + 2(-400)s
-800s = -6400
S = 6400/800
S = 8 m
so,