Q.The displacement of a body is proportional to t^3, where t is time elapsed. What is the nature of acceleration-time graph of the body ?
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Instantaneous Rate of Change
Imagine you're in a car watching the speedometer. It doesn't say "I travelled 60 km in the last hour" — it shows your speed right now, at this exact moment. That number, the one that changes every time you tap the brake or press the accelerator, is the instantaneous rate of change of your position.
The intuition: from average to instant
If you drive from Delhi to Agra (200 km) in 4 hours, your average speed is 50 km/h. But that tells you nothing about how fast you were going at 10:15 AM when you passed a particular toll booth. You might have been doing 80 km/h, or 20 km/h if there was traffic.
The average rate of change over a time interval [t1,t2] is:
Average speed=time takendistance travelled=t2−t1s(t2)−s(t1)
where s(t) is your position at time t.
To get the speed at a specific moment t=a, you'd want to look at smaller and smaller intervals around a. If you measure from t=a to t=a+h, where h is a tiny time difference:
Average speed over [a,a+h]=hs(a+h)−s(a)
As h gets smaller — say 0.1 seconds, then 0.01, then 0.0001 — this average speed gets closer and closer to a single number. That limiting number is the instantaneous rate of change at t=a.
This is the core idea: instantaneous rate of change = the limit of average rates of change as the interval shrinks to zero.
The precise definition
For any function y=f(x), the instantaneous rate of change at x=a is:
Instantaneous rate of change=limh→0hf(a+h)−f(a)
provided this limit exists. This limit is also called the derivative of f at a, denoted f′(a) or dxdyx=a.
f′(a)=limh→0hf(a+h)−f(a)
What it means geometrically
If you graph y=f(x), the average rate of change over [a,a+h] is the slope of the secant line through (a,f(a)) and (a+h,f(a+h)). As h→0, that secant line pivots and approaches a tangent line at x=a. The slope of that tangent line is exactly f′(a).
So instantaneous rate of change = slope of the tangent line.
A concrete example
Let f(x)=x2. Find the instantaneous rate of change at x=3.
First, the average rate over [3,3+h]:
hf(3+h)−f(3)=h(3+h)2−9=h9+6h+h2−9=h6h+h2=6+h
Now take the limit as h→0:
limh→0(6+h)=6
So at x=3, the function x2 is changing at a rate of 6 units per unit change in x. The tangent line at (3,9) has slope 6. …
Differentiate displacement twice with respect to time to get acceleration and see how it depends on t. …
If x∝t3, then a∝t, so the a-t graph is a straight line through the origin.
Let the displacement be:
x=kt3
where k is a constant.
Velocity is the first derivative of displacement with respect to time:
v=dtdx=3kt2
Acceleration is the derivative of velocity with respect to time:
a=dtdv=6kt
…
- CBSE 2024Set ANNUAL1 markMCQQ.The equation for displacement of a body is given by S = at + bt^2. The acceleration of the body is (A) a/b (B) 2b (C) a + b (D) 3b
›Reveal solutionSolution
For S=at+bt2, acceleration is constant and equal to 2b.
Velocity is the first derivative: v=dtdS=a+2bt.
…
- CBSE 2023Set annual1 markQ.If y = u^3 + 2u and u = x^2 + 5, find dy/dx.
›Reveal solutionSolution
Apply the chain rule: dy/dx = (dy/du) x (du/dx).
Given y = u^3 + 2u and u = x^2 + 5.
Step 1: Differentiate y with respect to u.
dy/du = 3u^2 + 2
Step 2: Differentiate u with respect to x.
du/dx = 2x
Step 3: Apply the chain rule.
dy/dx = (dy/du) x (du/dx) = (3u^2 + 2)(2x)
…
- CBSE 2021Set sz1 markQ.If y = cos^2 x, then dy/dx is .............
›Reveal solutionSolution
Using the chain rule on y = (cos x)^2 gives dy/dx = -2 sin x cos x, which can be written as -sin 2x.
Step 1: Write y = (cos x)^2 and let u = cos x, so y = u^2.
Step 2: Differentiate y with respect to u: dy/du = 2u.
Step 3: Differentiate u with respect to x: du/dx = -sin x.
Step 4: Apply the chain rule, dy/dx = (dy/du)(du/dx): …
- CBSE 2020Set annual1 markQ.If y = sqrt(x), find dy/dx.
›Reveal solutionSolution
For y = sqrt(x) = x^(1/2), the derivative is dy/dx = 1/(2 sqrt(x)), obtained by the power rule.
Given y = sqrt(x), rewrite it as y = x^(1/2).
Using the standard power-rule result d(x^n)/dx = n x^(n-1) with n = 1/2:
…
- CBSE 2020Set hz1 markQ.Find dy/dx, when y = 4x^3 + 7x^2 + 6x + 9
›Reveal solutionSolution
Differentiate each term of the polynomial using the power rule d/dx(x^n) = n x^(n-1).
Given y = 4x^3 + 7x^2 + 6x + 9.
Differentiate term by term:
d/dx(4x^3) = 4 * 3x^2 = 12x^2
d/dx(7x^2) = 7 * 2x = 14x
d/dx(6x) = 6 …
- CBSE 2019Set annual1 markQ.If y = 3x^2 + 4x + 5, find dy/dx, at x = 1.
›Reveal solutionSolution
Differentiate each term of y = 3x^2 + 4x + 5 using the power rule; dy/dx = 6x + 4, so at x = 1 the derivative is 10.
Step 1: Differentiate term by term.
d/dx(3x^2) = 3 * 2x = 6x
d/dx(4x) = 4
d/dx(5) = 0 (derivative of a constant is zero)
Step 2: Add the results. …
- CBSE 2018Set annual1 markQ.Given s = 4t^3 + 3, calculate ds/dt.
›Reveal solutionSolution
Differentiating s = 4t^3 + 3 with respect to t (using the power rule) gives ds/dt = 12t^2.
Given: s = 4t^3 + 3
Differentiate each term with respect to t. Using the power rule, d(t^n)/dt = n t^(n-1):
d(4t^3)/dt = 4 x 3 x t^(3-1) = 12 t^2
The derivative of the constant term (3) with respect to t is zero, since a constant does not change with time:
d(3)/dt = 0
Adding the two results:
…
- CBSE 2018Set annual1 markQ.Given s = 4t^3 + 3, calculate d^2s/dt^2.
›Reveal solutionSolution
Differentiating ds/dt = 12t^2 once more with respect to t gives the second derivative, d^2s/dt^2 = 24t.
From the previous part, ds/dt = 12t^2. The second derivative d^2s/dt^2 is obtained by differentiating this expression again with respect to t, using the power rule:
d^2s/dt^2 = d(12t^2)/dt = 12 x 2 x t^(2-1) = 24t
…
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