You visited us 1 times! Enjoying our articles? Unlock Full Access!

Motion Under Variable Acceleration

The position ...

Question

The position of a particle moving along the $x$ -axis is expressed as $x = a t^{3} + b t^{2} + c t + d$ . The initial acceleration of the particle is

6 a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 b

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(a + b)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a + c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 b$
We know that
$v = \frac{d x}{d t} = 3 a t^{2} + 2 b t + c$
Now for acceleration,
Let the acceleration be A,
$A = \frac{d v}{d t} = 6 a t + 2 b$
To find initial acceleration, we put $t = 0$ in above expression
At $t = 0$
$A = 2 b$

Suggest Corrections

Similar questions

x

t

x = A t^{3} + B t^{2} + C t + D

x

t

x = A t^{3} + B t^{2} + C t + D

X

X = A t^{3} + B t^{2} + C t + D .

A, B, C, D

1, 4, - 2

5

t = 0

t = 4 s

x = A t^{3} + B t^{3} + C t + D

A, B, C, D

1, 4, - 2, 5

t = 4 s

t

s = a t^{3} + b t^{3} + c t

a, b

c

Join BYJU'S Learning Program