Question

The motion of a particle is described by the equation X= at+bt2 where a= 15 cm S-1 and b = 3 cm S-2, Its acceleration at t= 3 sec will be:

Answer 1

Given :

• x = at + bt²
• a = 15
• b = 3

To Find :

• Acceleration when time is 3 seconds

Solution :

We're given that motion of a particle is described by the equation, x = at + bt². And the values of a and b are 15 and 3 respectively.

So, after putting values we get,

⇒x = 15t + 3t²

$\implies \sf{v \: = \: \dfrac{dx}{dt}} \\ \\ \implies \sf{v \: = \: \dfrac{d(15t \: + \: 3t^2)}{dt}} \\ \\ \implies \sf{v \: = \: 15 \: + \: 2(3t)} \\ \\ \implies \sf{v \: = \: 15 \: + \: 6t}$

$\therefore$ Velocity of the particle is $\sf{15 \: + \: 6t \: ms^{-1}}$

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Now, for finding acceleration we've to again differentiate the velocity :

$\implies \sf{a \: = \: \dfrac{dv}{dt}} \\ \\ \implies \sf{a \: = \: \dfrac{d(15 \: + \: 6t)}{dt}} \\ \\ \implies \sf{a \: = \: 0 \: + \: 6} \\ \\ \implies \sf{a \: = \: 6}$

$\therefore$ Particle is moving with constant acceleration of 6 m/s²