Rectilinear Kinematics (Constant Acceleration)

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Rectilinear Kinematics (Constant Acceleration)
Home Work 2-2
Q1) Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 10 m/s in 15 s, determine the constant deceleration of the car.
Q2) A car starts from rest and with constant acceleration achieves a velocity of 15 m/s when it travels a distance of 200 m. Determine the acceleration of the car and the time required.
Q3) A train starts from rest at a station and travels with a constant acceleration of 1 m/s². Determine the velocity of the train when t = 3s and the distance traveled during this time.
Q4) car is traveling at 15 m/s, when the traffic light 50 m ahead turns yellow. Determine the required constant deceleration of the car and the time needed to stop the car at the light.
Q5) Car A starts from rest at t = 0 and travels along a straight road with a constant acceleration of 6 ft/s² until it reaches a speed of 80 ft/s. Afterwards it maintains this speed. Also, when t=0, car B located 6000 ft down the road is traveling towards A at a constant speed of 60 ft/s. Determine the distance traveled by car A when they pass each other.

The Correct Answer and Explanation is :

Let’s go through each question step-by-step:

Q1) Determine the constant deceleration of the car.

  • Initial speed, vi=35 m/sv_i = 35 \, \text{m/s}
  • Final speed, vf=10 m/sv_f = 10 \, \text{m/s}
  • Time, t=15 st = 15 \, \text{s}
  • We can use the equation of motion: vf=vi+atv_f = v_i + a t Rearranging for acceleration (or deceleration): a=vf−vita = \frac{v_f – v_i}{t} Substituting values: a=10−3515=−2515=−1.67 m/s2a = \frac{10 – 35}{15} = \frac{-25}{15} = -1.67 \, \text{m/s}^2 Therefore, the constant deceleration is −1.67 m/s2-1.67 \, \text{m/s}^2.

Q2) Determine the acceleration and time required for the car to achieve 15 m/s after traveling 200 m.

  • Initial speed, vi=0 m/sv_i = 0 \, \text{m/s} (starts from rest)
  • Final speed, vf=15 m/sv_f = 15 \, \text{m/s}
  • Distance, d=200 md = 200 \, \text{m}
  • We can use the kinematic equation: vf2=vi2+2adv_f^2 = v_i^2 + 2 a d Substituting the values: 152=0+2a×20015^2 = 0 + 2 a \times 200 225=400a225 = 400a a=225400=0.5625 m/s2a = \frac{225}{400} = 0.5625 \, \text{m/s}^2 Now, using the equation vf=vi+atv_f = v_i + a t to find time tt: 15=0+0.5625t15 = 0 + 0.5625 t t=150.5625=26.67 st = \frac{15}{0.5625} = 26.67 \, \text{s} Therefore, the acceleration is 0.5625 m/s20.5625 \, \text{m/s}^2 and the time required is 26.67 s26.67 \, \text{s}.

Q3) Determine the velocity and distance traveled by the train after 3 seconds.

  • Initial velocity, vi=0 m/sv_i = 0 \, \text{m/s} (starts from rest)
  • Acceleration, a=1 m/s2a = 1 \, \text{m/s}^2
  • Time, t=3 st = 3 \, \text{s}
  • We can calculate the velocity using: v=vi+atv = v_i + a t v=0+1×3=3 m/sv = 0 + 1 \times 3 = 3 \, \text{m/s} For the distance, use: d=vit+12at2d = v_i t + \frac{1}{2} a t^2 d=0+12×1×32=4.5 md = 0 + \frac{1}{2} \times 1 \times 3^2 = 4.5 \, \text{m} Therefore, the velocity is 3 m/s3 \, \text{m/s} and the distance traveled is 4.5 m4.5 \, \text{m}.

Q4) Determine the deceleration and time needed for the car to stop at the light.

  • Initial speed, vi=15 m/sv_i = 15 \, \text{m/s}
  • Final speed, vf=0 m/sv_f = 0 \, \text{m/s}
  • Distance to the light, d=50 md = 50 \, \text{m}
  • We can use the equation: vf2=vi2+2adv_f^2 = v_i^2 + 2 a d Substituting values: 0=152+2a×500 = 15^2 + 2a \times 50 −225=100a-225 = 100a a=−225100=−2.25 m/s2a = \frac{-225}{100} = -2.25 \, \text{m/s}^2 Now, to find the time, use vf=vi+atv_f = v_i + a t: 0=15+(−2.25)t0 = 15 + (-2.25) t t=152.25=6.67 st = \frac{15}{2.25} = 6.67 \, \text{s} Therefore, the required deceleration is −2.25 m/s2-2.25 \, \text{m/s}^2 and the time is 6.67 s6.67 \, \text{s}.

Q5) Determine the distance traveled by car A when they pass each other.

  • Car A starts from rest with acceleration a=6 ft/s2a = 6 \, \text{ft/s}^2 until it reaches 80 ft/s80 \, \text{ft/s}.
  • Car B starts 6000 feet ahead of A and travels at 60 ft/s60 \, \text{ft/s}.
  • Car A’s speed at time tt is given by vA=6tv_A = 6t, and the distance it travels by time tt is dA=3t2d_A = 3t^2.
  • Car B’s position at time tt is dB=6000+60td_B = 6000 + 60t.

We need to solve for tt when dA=dBd_A = d_B: 3t2=6000+60t3t^2 = 6000 + 60t

Rearranging and solving: 3t2−60t−6000=03t^2 – 60t – 6000 = 0

Dividing by 3: t2−20t−2000=0t^2 – 20t – 2000 = 0

Solving using the quadratic formula: t=−(−20)±(−20)2−4(1)(−2000)2(1)=20±400+80002=20±84002t = \frac{-(-20) \pm \sqrt{(-20)^2 – 4(1)(-2000)}}{2(1)} = \frac{20 \pm \sqrt{400 + 8000}}{2} = \frac{20 \pm \sqrt{8400}}{2} t=20±91.652t = \frac{20 \pm 91.65}{2}

Taking the positive root: t=111.652=55.83 secondst = \frac{111.65}{2} = 55.83 \, \text{seconds}

Finally, the distance traveled by car A is: dA=3(55.83)2=9333.3 ftd_A = 3(55.83)^2 = 9333.3 \, \text{ft}

So, car A travels 9333.3 feet before they pass each other.

This covers all the required calculations and answers to the questions!

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