SATHEE: Motion in a Straight Line

Motion in a Straight Line - Result Question 24

25. The displacement of a particle is represented by the following equation : $s = 3 t^{3} + 7 t^{2} + 5 t + 8$ where $s$ is in metre and $t$ in second. The acceleration of the particle at $t = 1 s$ is

(a) $14 m / s^{2}$

(b) $18 m / s^{2}$

(d) zero

[2000]

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Answer:

Correct Answer: 25. (c)

Solution:

$s = 3 t^{3} + 7 t^{2} + 5 t + 8$

Velocity $= \frac{d s}{d t} = 9 t^{2} + 14 t + 5$

Acceleration $= \frac{d^{2} s}{d t^{2}} = 18 t + 14 ◻ = v_{max} (\frac{1}{α} + \frac{1}{β}) = v_{max} (\frac{α + β}{α β})$ Acceleration at $(t = 1 s)$ $= 18 \times 1 + 14 = 18 + 14 = 32 m / s^{2}$

Motion in a Straight Line - Result Question 24

25. The displacement of a particle is represented by the following equation : $s = 3 t^{3} + 7 t^{2} + 5 t + 8$ where $s$ is in metre and $t$ in second. The acceleration of the particle at $t = 1 s$ is

Answer:

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Motion in a Straight Line - Result Question 24

25. The displacement of a particle is represented by the following equation : s=3t3+7t2+5t+8 where s is in metre and t in second. The acceleration of the particle at t=1s is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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25. The displacement of a particle is represented by the following equation : $s = 3 t^{3} + 7 t^{2} + 5 t + 8$ where $s$ is in metre and $t$ in second. The acceleration of the particle at $t = 1 s$ is