A train is uniformly accelerated and passes successive kilometer stones with velocity of 18 kmph and 36 kmph respectively.Calculate the velocity, when it passes the thirdkilometer stone. Also find the time taken for each of these two intervals of one kilometer?
Answers
Answer:
Let us figure out the acceleration of the train. To do so, we will use the following kinematics equation:
v^2=v_0^2+2a(s-s_0)v
2
=v
0
2
+2a(s−s
0
)
(Where vv is final velocity, v_0v
0
is initial velocity, aa is constant acceleration, ss is final displacement, and s_0s
0
is initial displacement)
Let us substitute the values we know:
v^2=v_0^2+2a(s-s_0)v
2
=v
0
2
+2a(s−s
0
)
10^2=2^2+2a(1000-0)10
2
=2
2
+2a(1000−0)
(Here, the train travels 1000 m, starting at the origin of s_0=0s
0
=0 m. At the origin, it’s speed is 2 m/s, and at the 1000 m mark, the train has a speed of 10 m/s)
a=0.048a=0.048 m/s^2s
2
We can now find the speed of the train at the next marker. Note that the initial velocity of the train is now 10 m/s and we are trying to find the final velocity. As before, the distance between the markers is 1000 m. Using the same kinematics equation, we can write:
v^2=v_0^2+2a(s-s_0)v
2
=v
0
2
+2a(s−s
0
)
v^2=10^2+2(0.048)(1000-0)v
2
=10
2
+2(0.048)(1000−0)
v=14v=14 m/s
We can calculate the time it took for the train to reach the next marker using the following kinematics equation:
v=v_0+atv=v
0
+at
(Where vv is final velocity, v_0v
0
is initial velocity, aa is acceleration and tt is time)
Substitute the values we found:
v=v_0+atv=v
0
+at
14=2+(0.048)(t)14=2+(0.048)(t)
(Remember, we are looking for the time for the full journey. That means the initial velocity of the train was 2 m/s and final velocity was 14 m/s)
t=250t=250 s
Answer:
V = 13.22m/s
T₁ = 133.4
T₂=86.13
Explanation:
Given that:
Initial velocity,
U = 18km/hr = 18*5/18 = 5m/s.
Final velocity,
V = 36km/hr = 36*5/18 = 10m/s.
a = ? & V₃ = ?
V² - U² = 2as.
10² - 5² = 2 x a x 1000.
a = 0.375 ms⁻²
V₃² - U² = 2as.
V₃² = 2as + U²
V₃² = 2 x 0.375 x 2000 + 5².
V₃ =
V₃ = 13.22m/s
Time interval
V = U + a T₁
T₁ = V - U/a
T₁ = (10-5)/0.375 = 133.4 s
T₂ = (13.2-10)/0.375 = 86.13 s
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