18.A bus decreases its speed from 90km/hr to 36 km/hr in 10 seconds. Find the acceleration of the bus in m/s2.
1.5
-1.5
5.4
-5.4
Answers
Solution:
Initial velocity i.e. velocity before decreasing,v = 90 km/hr = =25m/s
Final velocity i.e. velocity after decreasing, u = 36 km/hr = = 10 m/s
Time taken to decrease velocity, t = 10 seconds
Let us apply the first law of equation of motion which is,
v = u + at
where,
v = Final velocity
u = Initial velocity
a = acceleration
t = time
v = u + at
10 = 25 + a(10)
10 - 25 = 10a
-15 = 10a
a = -15/10
a = -1.5 m/s²
The acceleration of the bus in m/s² is -1.5 m/s².
[Note => The acceleration is in negative as the bus speed is decreasing.]
Answer- The above question is from the chapter 'Kinematics'.
Some important terms and formulae:-
1. Velocity- It is the displacement per unit time.
S.I. Unit of Velocity- m/s
It is a vector quantity as it possesses magnitude and direction.
2. Acceleration- It is the rate of change of velocity.
S.I. Unit of Acceleration- m/s²
It is also a vector quantity.
Negative acceleration is called retardation.
3. Distance- It is the path length transversed by an object.
S.I. Unit of Distance- m
It is a scalar quantity.
4. Displacement- It is the shortest distance between the initial and final point.
S.I. Unit of Displacement- m
It is a vector quantity.
5. Equations for uniformly accelerated motion-
Let u = Initial velocity of a particle
v = Final velocity of a particle
t = Time taken
s = Distance travelled in the given time
a = Acceleration
1) v = u + at
2) s = at² + ut
3) v² - u² = 2as
Given question: A bus decreases its speed from 90km/hr to 36 km/hr in 10 seconds. Find the acceleration of the bus in m/s².
Answer: Initial velocity (u) = 90 km/hr
⇒ u = 90 × = 25 m/s
Final velocity (v) = 36 km/hr
⇒ v = 36 × = 10 m/s
Time (t) = 10 s
Using 1st equation of motion, v = u + at, we get,
10 = 25 + 10a
10 - 25 = 10a
-15 = 10a
⇒ a = -1.5 m/s²