4. A motorcycle moving with a speed of 5 m/s is subjected to an acceleration of 0.2 m/s2. Calculate the speed of
the motorcycle after 10 seconds, and the distance travelled in this time.
5. A bus running at a speed of 18 km/h is stopped in 2.5 seconds by applying brakes. Calculate the retardation
produced
train starting from rest moves with a uniform acceleration of 1.2 m/s2 for 5 minutes. Calculate the speed
Answers
Answer:
4) Given that
The initial speed of car=u= 5 m/s
Acceleration = a = 0.2 m/s
Time = t = 10 s
We need to calculate the distance
We know that
v = u + at
= 5 + 0.2×10
= 5+2
= 7 m/s
s = ut + (1/2)at²
= (5×10) + (1/2)×(2/10)×(10×10)
= 50 + 10
= 60 m
Final speed = 7 m/s
Distance travelled in 10 seconds = 60 m
5) Given that
Initial velocity = 18 km/h
We will express it in m/s
So Initial velocity = 18 km/h= 18×5/18 m/sec= 5 m/sec
The final velocity = zero
Acceleration = change in velocity/time
= (final velocity – initial velocity)/time
= (0 - 5)/2.5 = -2 m/s²
so, retardation = 2 m/s²
6) v = u + at
v = 0 + (1.2)(300) (5 minutes = 300 s)
v = 360 m/s
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Answer:
4. Initial velocity = u = 5 m/s
Acceleration = a = 0.2 m/s²
Time = t = 10 seconds
- Distance travelled by the bus is to be found (s)
Using second equation of motion:
S=ut+1/2 at²
S=5×10+1/2 × 0.2 × 10²
S=50+10
S= 60 metres
The motorcycle will cover a distance of 60 metres in the above given time
5. Initial velocity of the bus = u = 18 km/h = 18×5/18 = 5 m/s
Time taken for stopping the bus = t = 2.5 seconds
Final velocity of the bus = v = 0 m/s
- Acceleration is to be founhd here
Using first equation of motion:
V=u+at
0=5+2.5a
-5=2.5a
a = -2 m/s²
The acceleration of the bus is equal to -2 m/s²
6. Initial velocity = u = 0 m/s
Acceleration = a = 1.2 m/s²
Time = t = 5 minutes = 300 seconds
- Here final velocity (v) shall be found
Using first equation of motion:
V=u+at
V=0+1.2×300
V=360 m/s
The final velocity is equal to 360 m/s