A bus has a mass of 1400 kg is moving with a velocity of 36km/hr.If he driver applies the brakes to stop the bus within 4 sec.
1.Calculate retardation
2.calculate the distance covered by the bus by the time
3.calculate the force applied to stop the bus
Answers
Answer:
1.) Retardation = 2.5 m/s²
2.) Distance covered = 20 meters
3.) Force applied = -3500 N
Step by step explanations :
Given that,
A bus has a mass of 1400 kg is moving with a velocity of 36km/hr
here,
Mass of the bus = 1400 kg
Initial velocity of the bus = 36 km/h
= 36 × 1000 m/3600s
= 10 m/s
also given,
The driver applies the brakes to stop the bus within 4 sec.
so final velocity will be 0 km/h
because it will stop
time taken by bus to stop = 4 s
To find :
_____________________
Retardation:
We have,
Initial velocity(u) = 10 m/s
final velocity(v) = 0 m/s
time taken(t) = 4 s
by the equation of motion,
v = u + at
putting the values,
0 = 10 + a(4)
4a = -10
a = -10/4
a = -2.50 m/s²
Negation of acceleration shows Retardation.
so,
Retardation = 2.5 m/s²
now,
_______________________
The distance covered by the bus by the time:
Acceleration(a) = -2.5 m/s²
Initial velocity(u) = 10 m/s
time taken(t) = 4 s
by the equation of motion,
s = ut + ½ at²
again putting the values,
S = 10(4) + ½ (-2.5)(4)(4)
S = 40 - 20
S = 20 m
so,
Distance covered by time before coming to rest
= 20 meters
_____________________
the force applied to stop the bus
We have,
Mass of the bus(M) = 1400 kg
Acceleration(a) = -2.5 m/s²
Force = Ma
= 1400 × (-2.5)
= -3500 N
So,
Force applied by the brakes to stop the bus
Force applied by the brakes to stop the bus = -3500 N
___________
NOTE :
Negation of force shows the force applied I'm the opposite direction of motion.
__________________
1.) Retardation = 2.5 m/s²
2.) Distance covered = 20 meters
3.) Force applied = -3500 N
Answers—
1. ➡2.5 m/s²
2.➡20 m
3.➡ - 3500 N
--------------------------------------------------
Hᴇʀᴇ, Yᴏᴜʀs Sɪᴍᴘʟᴇ & Cᴀᴛᴄʜʏ Sᴏʟᴜᴛɪᴏɴs—
Given-
mass of bus = 1400 kg
Speed of bus initially = 36 km/h
Time taken to stop = 4 sec
1)
➡Retardaction means negative of acceleration
By the equation of motion,
v = u + at
putting the values,
0 = 10 + a(4)
4a = -10
a = -10/4
a = -2.50 m/s²
Therefore , Retardation = 2.5 m/s²
2)
distance covered by the bus
Acceleration (a) = -2.5 m/s²
Initial velocity(u) = 10 m/s
time taken(t) = 4 s
Again by the equation of motion,
s = ut + ½ at²
S = 10(4) + ½ (-2.5)(4)(4)
S = 40 - 20
S = 20 m
HENCE, Distance covered by time before coming to rest = 20 meters
3)
The force applied to stop
We know,
Mass of the bus(M) = 1400 kg
Acceleration(a) = -2.5 m/s²
Force = Ma
= 1400 × (-2.5)
= -3500 N