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Kinematics - Motion in a Straight Line

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Rest and Motion are relative terms. An object is in motion if its position changes with respect to a fixed frame of reference over time.

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Distance is the total path length covered, a scalar quantity. Displacement is the shortest straight-line distance between the initial and final positions, a vector quantity represented as Δx=xf−xi\Delta x = x_f - x_i.

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Average Speed is Total DistanceTotal Time\frac{\text{Total Distance}}{\text{Total Time}}, while Average Velocity is Net DisplacementTotal Time\frac{\text{Net Displacement}}{\text{Total Time}}.

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Instantaneous Velocity is defined as the rate of change of position at a specific instant: v=dxdtv = \frac{dx}{dt}.

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Acceleration is the rate of change of velocity. Instantaneous acceleration is a=dvdt=d2xdt2a = \frac{dv}{dt} = \frac{d^2x}{dt^2}. It can also be expressed as a=vdvdxa = v \frac{dv}{dx}.

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On a Position-Time (x−tx-t) graph, the slope represents the velocity.

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On a Velocity-Time (v−tv-t) graph, the slope represents the acceleration, and the area under the curve (considering sign) represents the displacement.

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Relative Velocity of body AA with respect to body BB in one dimension is given by vAB=vA−vBv_{AB} = v_A - v_B.

📐Formulae

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2=u2+2asv^2 = u^2 + 2as

sn=u+a2(2n−1)s_n = u + \frac{a}{2}(2n - 1) (Displacement in the nthn^{th} second)

vavg=u+v2v_{avg} = \frac{u + v}{2} (Only for uniform acceleration)

a=vdvdxa = v \frac{dv}{dx}

vrel=v1±v2v_{rel} = v_1 \pm v_2

💡Examples

Problem 1:

A car starts from rest and accelerates uniformly at 2 m/s22\text{ m/s}^2 for 10 s10\text{ s}. It then travels at a constant velocity for 20 s20\text{ s} and finally comes to rest in 5 s5\text{ s} under uniform retardation. Find the total distance covered.

Solution:

  1. Phase 1 (Acceleration): u=0,a=2,t=10u = 0, a = 2, t = 10. v=u+at=0+2(10)=20 m/sv = u + at = 0 + 2(10) = 20\text{ m/s}. Distance s1=ut+12at2=0+12(2)(10)2=100 ms_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2)(10)^2 = 100\text{ m}.
  2. Phase 2 (Constant Velocity): v=20,t=20v = 20, t = 20. Distance s2=v×t=20×20=400 ms_2 = v \times t = 20 \times 20 = 400\text{ m}.
  3. Phase 3 (Retardation): u=20,v=0,t=5u = 20, v = 0, t = 5. Distance s3=u+v2×t=20+02×5=50 ms_3 = \frac{u+v}{2} \times t = \frac{20+0}{2} \times 5 = 50\text{ m}. Total distance S=s1+s2+s3=100+400+50=550 mS = s_1 + s_2 + s_3 = 100 + 400 + 50 = 550\text{ m}.

Explanation:

The motion is divided into three segments. We use the equations of motion for the accelerated segments and simple velocity-time product for the uniform motion segment.

Problem 2:

The displacement of a particle moving along the x-axis is given by x=18t+5t2x = 18t + 5t^2, where xx is in meters and tt is in seconds. Calculate the instantaneous velocity and acceleration at t=2 st = 2\text{ s}.

Solution:

Displacement x=18t+5t2x = 18t + 5t^2. Velocity v=dxdt=ddt(18t+5t2)=18+10tv = \frac{dx}{dt} = \frac{d}{dt}(18t + 5t^2) = 18 + 10t. At t=2 st = 2\text{ s}, v=18+10(2)=38 m/sv = 18 + 10(2) = 38\text{ m/s}. Acceleration a=dvdt=ddt(18+10t)=10 m/s2a = \frac{dv}{dt} = \frac{d}{dt}(18 + 10t) = 10\text{ m/s}^2.

Explanation:

Calculus is used here: velocity is the first derivative of displacement with respect to time, and acceleration is the derivative of velocity.

Problem 3:

A ball is thrown vertically upwards with a velocity of 20 m/s20\text{ m/s} from the top of a tower 25 m25\text{ m} high. How long will it take for the ball to hit the ground? (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

Taking upward direction as positive: u=+20 m/su = +20\text{ m/s}, a=−g=−10 m/s2a = -g = -10\text{ m/s}^2, total displacement s=−25 ms = -25\text{ m} (since it ends up below the starting point). Using s=ut+12at2s = ut + \frac{1}{2}at^2: −25=20t+12(−10)t2-25 = 20t + \frac{1}{2}(-10)t^2 −25=20t−5t2-25 = 20t - 5t^2 5t2−20t−25=05t^2 - 20t - 25 = 0 t2−4t−5=0t^2 - 4t - 5 = 0 (t−5)(t+1)=0(t-5)(t+1) = 0. Since time cannot be negative, t=5 st = 5\text{ s}.

Explanation:

By using the displacement as −25 m-25\text{ m} in the equation of motion, we account for the entire trajectory (upward and then downward) in a single step.