Soru

Zorluk: KolayKinematics and Linear Motion

A body accelerates uniformly from rest at a rate of 4 m/s24\text{ m/s}^2 for 6 s6\text{ s}. What is the distance covered by the body during this time interval?

  1. 72 m72\text{ m}Cevap
  2. B
    144 m144\text{ m}
  3. C
    24 m24\text{ m}
  4. D
    12 m12\text{ m}

Cevap

The distance covered by the body is 72 m72\text{ m}.
Applying the equation of motion s=ut+12at2s = ut + \frac{1}{2}at^2 with u=0 m/su = 0\text{ m/s}, a=4 m/s2a = 4\text{ m/s}^2, and t=6 st = 6\text{ s} yields s=0+12(4)(62)=72 ms = 0 + \frac{1}{2}(4)(6^2) = 72\text{ m}.

Adım Adım Çözüm

1
Identify the given kinematic parameters
Initial velocity u=0 m/su = 0\text{ m/s}, acceleration a=4 m/s2a = 4\text{ m/s}^2, and time interval t=6 st = 6\text{ s}.
Since the body starts from rest, its initial velocity is zero.
2
Select and set up the equation of motion for displacement
s=ut+12at2=(0)(6)+12(4)(6)2s = ut + \frac{1}{2}at^2 = (0)(6) + \frac{1}{2}(4)(6)^2
This formula directly relates displacement to initial velocity, acceleration, and time under constant acceleration.
3
Calculate the total distance
s=12×4×36=72 ms = \frac{1}{2} \times 4 \times 36 = 72\text{ m}
Squaring 6 s6\text{ s} yields 36 s236\text{ s}^2, and multiplying by 2 m/s22\text{ m/s}^2 gives 72 m72\text{ m}.

Anahtar Kavram

Linear motion under uniform acceleration
Tahmini Süre:45s
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