Motion in a straight line practice · Easy

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Application +4 | -1

The displacement of a particle is given by

y=a+b t+c t^2-d t^4

. The initial velocity and acceleration are respectively

View answer and explanation

Correct option: C

b, 2c

Option C (b, 2c) is correct because the initial velocity is b and the initial acceleration is 2c, obtained by differentiating the displacement equation.

Solution approach

Correct Answer + Core Reason: b, 2c. The initial velocity is b and the initial acceleration is 2c.

Step-by-Step Solution:

Step 1: Given displacement equation: $y = a + bt + ct^2 - dt^4$

Step 2: Find velocity by differentiating displacement with respect to time: $v = \frac{{dy}}{{dt}} = b + 2ct - 4dt^3$

Step 3: Find acceleration by differentiating velocity with respect to time: $a = \frac{{dv}}{{dt}} = 2c - 12dt^2$

Step 4: Find initial velocity by setting t=0 in the velocity equation: $v_{initial} = b + 2c(0) - 4d(0)^3 = b$

Step 5: Find initial acceleration by setting t=0 in the acceleration equation: $a_{initial} = 2c - 12d(0)^2 = 2c$

Step 6: Final Answer: Initial velocity is b and initial acceleration is 2c.

Why other options are wrong:
- Option A: Incorrect because the initial acceleration is not -4d.
- Option B: Incorrect because the initial velocity is not -b.
- Option D: Incorrect because the initial velocity is not 2c and the initial acceleration is not -4d.

Key Concepts:
- Differentiation
- Velocity and acceleration as derivatives of displacement

Memory Aids:
- Remember that velocity is the first derivative of displacement, and acceleration is the first derivative of velocity (or the second derivative of displacement).

Quick Tips:
- Differentiate the displacement equation to find velocity and acceleration, then substitute t=0 to find the initial values.

Revision notes and key takeaways

Review MOTION IN A STRAIGHT LINE in your textbook or class notes, then use this set to check your understanding.

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