Question

The equation of motion of a particle starting from rest along a straight line is x=t3-3t2+5 the ratio of the acceleration after 5 sec and 3 sec will be

Answer 1

$x = {t}^{3} - 3 {t}^{2} + 5$
$v = dx \div dt = 3 {t}^{2} - 6t$
$a = {d}^{2} x \div d {t }^{2} = 6t - 6$
aceleration at t=5 is 24
accelertion at t=3 is 12
ratio is 24/12=2:1

Answer 2

Answer:

Ratio = 2:1

Step-by-step explanation:

Given : The equation of motion of a particle starting from rest along a straight line is $x=t^3-3t^2+5$

To find : The ratio of the acceleration after 5 sec and 3 sec will be?

Solution :

The equation of motion of a particle starting from rest along a straight line is $x=t^3-3t^2+5$

Velocity is defined by first derivative w.r.t t

$v = \frac{dx}{dt} = 3 {t}^{2} - 6t$

Acceleration is defined by second derivative w.r.t t

$a =\frac{d^2x}{dt^2} = 6t - 6$

Now, Substituting t= 5 and 3 to find ratio,

Acceleration at t=5

$a= 6(5) - 6=30-6=24$

Acceleration at t=3

$a= 6(3) - 6=18-6=12$

The ratio of their acceleration is

$\frac{24}{12}=\frac{2}{1}$

Ratio = 2:1