Question

A particle moves along a straight line such that its
displacement at any time t is given by
$s = 3 {t}^{3} + 7 {t}^{2} + 14t + 5$



The acceleration of the particle at t=1s is :
(a) 18 m/s
(b)32 m/s
(c) 29 m/s?
(d) 24 m/s​

Answer 1

Answer:-

a = 32 m/s²

Option → B

Given :-

$S = 3t^3 +7t^2 +14 t +5$

To find :-

The acceleration of the particle at

t = 1 s.

Solution:-

→$S = 3t^3 +7t^2 +14 t +5$

• Differentiate with respect to time.

→$\dfrac{ds}{dt}= \dfrac{d(3t^3 +7t^2 +14 t +5)}{dt}$

→$\dfrac{ds}{dt}= 3\dfrac{d(t^3)}{dt}+ 7\dfrac{d(t^2)}{dt}+14\dfrac{dt}{dt}+ \dfrac{d(5)}{dt}$

→$v = 3 \times 3 t^2 + 7 \times 2 t + 14 + 0$

→$v = 9t^2 +14t +14$

• Differentiate again with respect to time.

→$\dfrac{dv}{dt} = \dfrac{d(9t^2 +14t +14)}{dt}$

→$\dfrac{dv}{dt}=9 \dfrac{d(t^2)}{dt} + 14 \dfrac{dt}{dt}+ \dfrac{d(14)}{dt}$

→$a = 9 \times 2t + 14 \times 1 +0$

→$a = 18t +14$

• at t = 1s

The acceleration of the particle will be :-

→$a = 18 \times 1 + 14$

→$a = 18 +14$

→$a = 32 m/s^2$

hence,

The acceleration will be 32 m/s².

Answer 2

Answer :

Given -

• s = 3t³ + 7t² + 14t + 5

To Find -

• Acceleration at t = 1 second ?

Solution -

Firstly let us differentiate to find Instantaneous velocity :

⇒ v = ds/dt

⇒ v = d(3t³ + 7t² + 14t + 5)/dt

⇒ v = 3*3t² + 7*2t + 14 + 0

⇒ v = 9t² + 14t + 14

Now, let us differentiate again to find Instantaneous acceleration :

⇒ a = dv/dt

⇒ a = d(9t² + 14t + 14)/dt

⇒ a = 2*9t + 14 + 0

⇒ a = 18t + 14

Now, Instantaneous acceleration at t = 1 second :

⇒ 18t + 14

⇒ 18*1 + 14

⇒ 18 + 14

⇒ 32 m/s²

Hence, Instantaneous acceleration at t = 1 second is 32 m/s².