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Difficulty: HardKinematics and Linear Motion

An electric train starts from rest and accelerates uniformly at 2 m/s22\text{ m/s}^2 for a time interval tt. It then maintains the maximum velocity attained for a time interval of tt. Finally, it decelerates uniformly at 4 m/s24\text{ m/s}^2 until it comes to a complete stop. If the total distance covered during the entire journey is 350 m350\text{ m}, what is the value of tt?

  1. A
    5 s5\text{ s}
  2. B
    7 s7\text{ s}
  3. 10 s10\text{ s}Answer
  4. D
    14 s14\text{ s}

Answer

10 s10\text{ s}
The motion consists of three stages: acceleration from rest (s1=t2s_1 = t^2), uniform velocity (s2=2t2s_2 = 2t^2), and deceleration to rest (s3=0.5t2s_3 = 0.5t^2). Adding these gives a total distance of S=3.5t2S = 3.5t^2. Setting 3.5t2=350 m3.5t^2 = 350\text{ m} yields t2=100t^2 = 100, so t=10 st = 10\text{ s}.

Step-by-Step Solution

1
Analyze Stage 1 (Acceleration phase)
Maximum velocity v=2tv = 2t, displacement s1=12(2)t2=t2s_1 = \frac{1}{2}(2)t^2 = t^2
Starting from rest (u=0u = 0) with acceleration a1=2 m/s2a_1 = 2\text{ m/s}^2 for duration tt, v=u+a1t=2tv = u + a_1 t = 2t and s1=12a1t2=t2s_1 = \frac{1}{2} a_1 t^2 = t^2.
2
Analyze Stage 2 (Constant velocity phase)
Displacement s2=(2t)(t)=2t2s_2 = (2t)(t) = 2t^2
The train moves at constant velocity v=2tv = 2t for duration tt, so distance is velocity multiplied by time.
3
Analyze Stage 3 (Deceleration phase)
Deceleration time t3=0.5tt_3 = 0.5t, displacement s3=0.5t2s_3 = 0.5t^2
Decelerating from v=2tv = 2t to 00 at a2=4 m/s2a_2 = 4\text{ m/s}^2 takes time t3=va2=2t4=0.5tt_3 = \frac{v}{a_2} = \frac{2t}{4} = 0.5t. Distance s3=12vt3=12(2t)(0.5t)=0.5t2s_3 = \frac{1}{2} v t_3 = \frac{1}{2} (2t)(0.5t) = 0.5t^2.
4
Calculate total displacement and solve for tt
Total distance S=3.5t2=350  ⟹  t2=100  ⟹  t=10 sS = 3.5t^2 = 350 \implies t^2 = 100 \implies t = 10\text{ s}
Summing the displacements: S=t2+2t2+0.5t2=3.5t2S = t^2 + 2t^2 + 0.5t^2 = 3.5t^2. Equating to 350 m350\text{ m} gives 3.5t2=3503.5t^2 = 350, so t2=100t^2 = 100, giving t=10 st = 10\text{ s}.

Key Concept

Multi-stage linear motion and displacement calculation
Estimated Time:2m 0s
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Topics

Kinematics and Linear Motion

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