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Motion in a Straight Line - Acceleration

Grade 11CBSEPhysics

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

🔑Concepts

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Acceleration is defined as the rate of change of velocity with respect to time. It is a vector quantity, meaning it has both magnitude and direction. Its SI unit is m/s2m/s^2 and its dimensional formula is [M0L1T−2][M^0 L^1 T^{-2}].

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Average acceleration is the change in velocity divided by the time interval taken for that change: aavg=ΔvΔt=v2−v1t2−t1a_{avg} = \frac{\Delta v}{\Delta t} = \frac{v_2 - v_1}{t_2 - t_1}.

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Instantaneous acceleration is the acceleration at a specific instant of time, defined as the derivative of velocity with respect to time: a=dvdt=d2xdt2a = \frac{dv}{dt} = \frac{d^2x}{dt^2}. It can also be written in terms of displacement as a=vdvdxa = v \frac{dv}{dx}.

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Uniform acceleration occurs when the velocity of an object changes by equal amounts in equal intervals of time. For such motion, the kinematic equations are valid.

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The slope of a velocity-time (v−tv-t) graph represents the acceleration of the object. A constant slope indicates uniform acceleration, while a changing slope indicates non-uniform acceleration.

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Retardation or deceleration is acceleration in the direction opposite to the velocity, leading to a decrease in the speed of the object. It is mathematically represented as a negative value if the direction of motion is positive.

📐Formulae

a=v−uta = \frac{v - u}{t}

a=dvdt=d2xdt2a = \frac{dv}{dt} = \frac{d^2x}{dt^2}

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

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)

💡Examples

Problem 1:

A car moving at a speed of 30 m/s30\,m/s is brought to a halt in 5 s5\,s by applying brakes. Calculate the acceleration of the car.

Solution:

Given: initial velocity u=30 m/su = 30\,m/s, final velocity v=0 m/sv = 0\,m/s, and time t=5 st = 5\,s. Using the formula a=v−uta = \frac{v - u}{t}:

a=0−305a = \frac{0 - 30}{5} a=−6 m/s2a = -6\,m/s^2

Explanation:

The negative sign indicates that the car is decelerating (retardation). The magnitude of deceleration is 6 m/s26\,m/s^2.

Problem 2:

The velocity of a particle is given by the equation v=2t2+5t m/sv = 2t^2 + 5t\,m/s. Find the instantaneous acceleration at t=3 st = 3\,s.

Solution:

To find acceleration, we differentiate velocity with respect to time: a=dvdt=ddt(2t2+5t)a = \frac{dv}{dt} = \frac{d}{dt}(2t^2 + 5t) a=4t+5a = 4t + 5

Substituting t=3t = 3: a=4(3)+5a = 4(3) + 5 a=12+5=17 m/s2a = 12 + 5 = 17\,m/s^2

Explanation:

Acceleration is the first derivative of velocity. By applying the power rule ddt(tn)=ntn−1\frac{d}{dt}(t^n) = nt^{n-1}, we find the acceleration function and evaluate it at the given time.

Problem 3:

A bus reduces its speed from 85 m/s85\,m/s to 42 m/s42\,m/s. Calculate the total change in velocity.

Solution:

Change in velocity Δv=vfinal−vinitial\Delta v = v_{final} - v_{initial}. Using vertical subtraction: 85−4243\begin{array}{r} 85 \\ -42 \\ \hline 43 \end{array}

Δv=−43 m/s\Delta v = -43\,m/s (since it is a reduction).

Explanation:

The change is calculated by subtracting the initial velocity from the final velocity. The arithmetic shows a magnitude of 43 m/s43\,m/s.