Question

The acceleration of a particle is given by a=t³-3t²+5, where a is in m/s² and t is in seconds. At t=1s, the displacement and velocity are 8.30m and 6.25m/s, respectively. Calculate the displacement and velocity at t=2s.​

Answer 1

EXPLANATION.

The acceleration of a particle is given

=> a = t³ - 3t² + 5 , where a =m/s²

t is in second.

At t = 1 sec the displacement and velocity

are 8.30 m and 6.25 m/s

Calculate the displacement and

velocity at t = 2 seconds.

According to the question,

=> a = t³ - 3t² + 5

$= > \: \: \bold{v \: \: = \int \: adt}$

$\bold{= > \: \: v \: = \int \: (t {}^{3} - 3t {}^{2} + 5)dt }$

=> at = t = 1 seconds and v = 6.25 m/s

$\bold{= > \: \: v \: = \frac{t {}^{4} }{4} - \frac{3t {}^{3} }{3} + 5t \: + c_{1}}$

put the value we get,

=> 6.25 = 1/4 - 1 + 5 + c¹

=> c¹ = 2

$\bold{v \: = \frac{t {}^{4} }{4} - t {}^{3} + 5t + 2 }$

$\bold{x \: = \int \: vdt}$

$\bold{x \: = \frac{t {}^{5} }{20} - \frac{1}{4} + \frac{5}{2} + + c_{2}}$

=> c² = 4

$\bold{x \: = \frac{t {}^{5} }{20} - \frac{t {}^{4} }{4} + \frac{5t {}^{2} }{2} + 2t \: + 4 }$

put t = 2 seconds we get,

=> v = (2)^4 / 4 - 2³ + 5 X 2 + 2

=> v = 8 m/s

x = (2)^5/20 - (2)^4/4 + 5/2(2)²+2X2 + 4

=> x = 15.6 m