The relation between time t and distance x is x = at2+bt3+c, where a a

1.	The relation between time t and distance x is x = at2+bt3+c, where a and b are constants. The acceleration at t = 2 s.
Answer» Given : ➳ Relation b/w time and distance has been provided. $\bigstar\:\underline{\boxed{\tt{x=at^2+bt^3+c}}}$ To Find : ➨ Acc. at t = 2 s Solution : ✭ Instantaneous velocity : $\dashrightarrow\tt\:v=\dfrac{dx}{dt}\\ \\ \dashrightarrow\tt\:v=\dfrac{d(at^2+bt^3+c)}{dt}\\ \\ \dashrightarrow\tt\:v=2at+3bt^2+0\\ \\ \dashrightarrow\bf\:v=2at+3bt^2$ ✭ Instantaneous ACCELERATION : $\longrightarrow\tt\:a=\dfrac{dv}{dt}\\ \\ \longrightarrow\tt\:a=\dfrac{d(2at+3bt^2)}{dt}\\ \\ \longrightarrow\bf\:a=2a+6bt$ PUTTING t = 2, we GET $\longrightarrow\tt\:a=2a+6b(2)\\ \\ \longrightarrow\underline{\boxed{\bf{a=2(a+6b)}}}$

The relation between time t and distance x is x = at2+bt3+c, where a and b are constants. The acceleration at t = 2 s.

Answer»

Given :

➳ Relation b/w time and distance has been provided.

$\bigstar\:\underline{\boxed{\tt{x=at^2+bt^3+c}}}$

To Find :

➨ Acc. at t = 2 s

Solution :

✭ Instantaneous velocity :

$\dashrightarrow\tt\:v=\dfrac{dx}{dt}\\ \\ \dashrightarrow\tt\:v=\dfrac{d(at^2+bt^3+c)}{dt}\\ \\ \dashrightarrow\tt\:v=2at+3bt^2+0\\ \\ \dashrightarrow\bf\:v=2at+3bt^2$