Question

The position of an object moving along
x-axis is given by x = a + bt^2? where a = 8.5 m,
b 2.5 ms-2 and t is measured in seconds.
Then which of the following is true ?
1) Velocity at t=2 sec is zero
2) average velocity between t=2, t=4 sec is
15 m/s.
3) velocity at t = 4 sec is 10 m/s
4) all the above are true​
pls help

Answer 1

Given :

▪ Position-time equation of an object moving along x-axis has been provided.

$\bigstar\:\underline{\boxed{\bf{\red{x=a+bt^2}}}}$

To Find :

▪ Velocity at t = 2s

▪ Velocity at t = 4s

▪ Average velocity b/w t = 2s and t = 4s.

SoluTioN :

Instantaneous velocity is given by

$\bigstar\:\underline{\boxed{\bf{\pink{v=lim(\Delta t\to 0)\:\dfrac{\Delta r}{\Delta t}=\dfrac{dr}{dt}}}}}$

$\dashrightarrow\sf\:v=\dfrac{dx}{dt}\\ \\ \dashrightarrow\sf\:v=\dfrac{d(a+bt^2)}{dt}\\ \\ \dashrightarrow\bf\:v=2bt$

❇ Option - A :

$\implies\sf\:v=2bt\\ \\ \implies\sf\:v=2(2.5)(2)\\ \\ \implies\underline{\boxed{\bf{\blue{v_1=10\:mps}}}}$

❇ Option - C :

$\mapsto\sf\:v=2bt\\ \\ \mapsto\sf\:v=2(2.5)(4)\\ \\ \mapsto\underline{\boxed{\bf{\green{v_2=20\:mps}}}}$

❇ Option - B :

$\Rightarrow\sf\:v_{av}=\dfrac{v_1+v_2}{2}\\ \\ \Rightarrow\sf\:v_{av}=\dfrac{10+20}{2}\\ \\ \Rightarrow\underline{\boxed{\bf{\purple{v_{av}=15\:mps}}}}$

____________________________________

Option-A : Incorrect

Option-B : Correct

Option-C : Incorrect

Answer 2

Answer :

Given -

• Position of an object = x = a + bt²
• a = 8.5 m
• b = 2.5 m/s²
• t is in seconds

Solution -

Instantaneous velocity = dx/dt.

⇒ d(a + bt²)/dt

⇒ 2bt

Now, velocity at t = 2 seconds :

⇒ v1 = 2bt

⇒ v1 = 2*2.5*2

⇒ v1 = 10 m/s

Also, velocity at t = 4 seconds :

⇒ v2 = 2bt

⇒ v2 = 2*2.5*4

⇒ v2 = 20 m/s

Average speed : (v1 + v2)/2

⇒ (10 + 20)/2

⇒ 30/2

⇒ 15 m/s

Final Answer :

• Option A is wrong.
• Option B is correct.
• Option C is wrong.
• Option D is wrong.