3. A train starting from a railway
station and moving with uniform
acceleration attains a speed
40 kmh in 10 minutes. Find its
acceleration.
Answers
Answer:
Known Terms:-
u = Initial speed
v = Final speed
a = Acceleration
t = Time
Given:-
Initial speed of the train, u = 0 km (Because train starts from the rest.
Final speed of the train, v = 40 km/h = \frac{5}{18}
18
5
× 40 = 11.11 m/s.
Time taken to attain the speed = 10 minutes = 10 × 60 = 600 sec.
To Calculate:-
Acceleration of the train.
Formula to be used:-
⇒ a=\frac{v-u}{t}a=
t
v−u
Solution:-
Putting all the values, we get
⇒ a=\frac{v-u}{t}a=
t
v−u
⇒ a=\frac{11.11-0}{600}a=
600
11.11−0
⇒ a=\frac{11.11}{600}a=
600
11.11
⇒ a=0.0185 \: m/s^2a=0.0185m/s
2
Hence, Acceleration of the train is 0.0185 m/s².
Read more on Brainly.in - https://brainly.in/question/18865598#readmoreAnswer:
Known Terms:-
u = Initial speed
v = Final speed
a = Acceleration
t = Time
Given:-
Initial speed of the train, u = 0 km (Because train starts from the rest.
Final speed of the train, v = 40 km/h = \frac{5}{18}
18
5
× 40 = 11.11 m/s.
Time taken to attain the speed = 10 minutes = 10 × 60 = 600 sec.
To Calculate:-
Acceleration of the train.
Formula to be used:-
⇒ a=\frac{v-u}{t}a=
t
v−u
Solution:-
Putting all the values, we get
⇒ a=\frac{v-u}{t}a=
t
v−u
⇒ a=\frac{11.11-0}{600}a=
600
11.11−0
⇒ a=\frac{11.11}{600}a=
600
11.11
⇒ a=0.0185 \: m/s^2a=0.0185m/s
2
Hence, Acceleration of the train is 0.0185 m/s².
Read more on Brainly.in - https://brainly.in/question/18865598#readmoreAnswer:
Known Terms:-
u = Initial speed
v = Final speed
a = Acceleration
t = Time
Given:-
Initial speed of the train, u = 0 km (Because train starts from the rest.
Final speed of the train, v = 40 km/h = \frac{5}{18}
18
5
× 40 = 11.11 m/s.
Time taken to attain the speed = 10 minutes = 10 × 60 = 600 sec.
To Calculate:-
Acceleration of the train.
Formula to be used:-
⇒ a=\frac{v-u}{t}a=
t
v−u
Solution:-
Putting all the values, we get
⇒ a=\frac{v-u}{t}a=
t
v−u
⇒ a=\frac{11.11-0}{600}a=
600
11.11−0
⇒ a=\frac{11.11}{600}a=
600
11.11
⇒ a=0.0185 \: m/s^2a=0.0185m/s
2
Hence, Acceleration of the train is 0.0185 m/s².
Answer:
Known Terms:-
u = Initial speed
v = Final speed
a = Acceleration
t = Time
Given:-
Initial speed of the train, u = 0 km (Because train starts from the rest.
Final speed of the train, v = 40 km/h = \frac{5}{18}
18
5
× 40 = 11.11 m/s.
Time taken to attain the speed = 10 minutes = 10 × 60 = 600 sec.
To Calculate:-
Acceleration of the train.
Formula to be used:-
⇒ a=\frac{v-u}{t}a=
t
v−u
Solution:-
Putting all the values, we get
⇒ a=\frac{v-u}{t}a=
t
v−u
⇒ a=\frac{11.11-0}{600}a=
600
11.11−0
⇒ a=\frac{11.11}{600}a=
600
11.11
⇒ a=0.0185 \: m/s^2a=0.0185m/s
2
Hence, Acceleration of the train is 0.0185 m/s².