Question

A particle moves in a straight line so that it covers a distance S=at^3+bt+5 meter in t second. If it's acceleration after 4 seconds is 48m/s^2,then a is equal to

Answer 1

this is the explanation and answer. we have differentiate on both sides with respect to t in first two steps.

Answer 2

Answer:

A particle moves in a straight line so that it covers a distance S=at^3+bt+5 meter in t second. If it's acceleration after 4 seconds is 48 m/$s^{2}$,then a is equal to 2.

Step-by-step explanation:

Given, the equation of distance covered by the particle is $S = at^{3} +bt+5$, where t is in seconds.

Acceleration is defined as the rate of change of velocity and velocity is defined as the rate of change of displacement/distance.

Hence to convert the distance equation into acceleration equation, it should be differentiated with respect to time twice.

Velocity = $\frac{dS}{dt} = \frac{d(at^{3}+bt+5) }{dt}$

Since $\frac{d(x^{n} )}{dx} = n*x^{n-1} , \frac{d(constant)}{dt} = 0$,    

Velocity  = $3at^{2}+b$

Acceleration = $\frac{d(velocity)}{dt} = \frac{d(3at^{2} )}{dt}$

Acceleration = 6at

Acceleration after 4 seconds is 48 m/$s^{2}$.

48 = 6*a*4

a = 2

Hence, the value of a is 2.

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