A train starts from rest and accelerates uniformly at 100 m/min2 for 10 minutes. Find the velocity acquired by the train. It then maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity-time graph and use it to find: i) the retardation in the last 5 minutes ii) total distance travelled, and iii) the average velocity of the train
Answers
Answer:
Let ‘u’ be initial velocity.
Let ‘v’ be final velocity.
The final velocity of the train is:
\bold{v = u + at}v=u+at
\bold{v = 0 + (100 \times 10)}v=0+(100×10)
v = 1000 m/s
So, this is the maximum velocity of the train.
The distance covered is given by the formula:
\bold{S = \frac{1}{2}at^2}S=
2
1
at
2
Distance covered during uniform acceleration:
\bold{S_{1} = \frac{1}{2}at^{2}_{1}}S
1
=
2
1
at
1
2
\bold{S_{1} = \frac{1}{2} \times 100 \times 10^{2}}S
1
=
2
1
×100×10
2
\bold{S_{1} = 5000 \ m = 5 \ km}S
1
=5000 m=5 km
Distance covered during constant velocity:
\bold{S_{2} = v \times t_{2}}S
2
=v×t
2
\bold{S_{2} = 1000 \times 20}S
2
=1000×20
\bold{S_{2} = 20000 \ m = 20 \ km}S
2
=20000 m=20 km
Distance covered during retardation:
Retardation = \bold{200 \ m/min^{2}}200 m/min
2
\bold{S_{3} = \frac{1}{2}at^{2}_{3}}S
3
=
2
1
at
3
2
\bold{S_{3} = \frac{1}{2} \times 200 \times 5^{2}}S
3
=
2
1
×200×5
2
\bold{S_{3} = 2500 \ m = 2.5 \ km}S
3
=2500 m=2.5 km
Now, the total distance is:
\bold{S = S_{1} + S_{2} + S_{3}}S=S
1
+S
2
+S
3
\bold{S = 5 + 20 + 2.5}S=5+20+2.5
\bold{\therefore S = 27.5 \ km}∴S=27.5 km
Explanation:
Total distance covered(s) = Distance during acceleration(s
1
) + distance during uniform motion(s
2
) + distance during retardation(s
3
)
For acceleration,
v=u+a∗t
v
max
=2×10=20 m/s
s=ut+(1/2)at
2
s
1
=(1/2)×2×10
2
=100 m
For uniform motion,
s=vt
s
2
=20×200
=4000 m
During retardation,
v=u+at
0=20+50t
=−0.4 m/s
2
s
3
=ut+(1/2)at
2
=20×50+(1/2)×(−0.4)×50
2
=500 m
Total distance travelled, s=s1+s2+s3=100+4000+500=4600 m
Total time taken, t=t
1
+t
2
+t
3
=260 s
Average velocity, v
avg
=
Total Time
Total Distance
=4600/260=17.69 m/s