The displacement is given by x = 2t^2 + t + 5 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the displacement equation: $x = 2t^2 + t + 5$.Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt}

Vedclass · Accepted Answer

$4 \ m \cdot s^{-2}$

Vedclass · Answer

(A) Given the displacement equation: $x = 2t^2 + t + 5$.Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(2t^2 + t + 5) = 4t + 1$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(4t + 1) = 4 \ m \cdot s^{-2}$.Since the acceleration is constant,the acceleration at $t = 2 \ s$ is $4 \ m \cdot s^{-2}$.

The displacement is given by $x = 2t^2 + t + 5$. The acceleration at $t = 2 \ s$ is:

Similar Questions

$A$ body starts from the origin and moves along the $x$-axis such that its position at any instant is $x = 4t^2 - 12t$,where $t$ is in seconds and $x$ is in meters. What is the acceleration of the particle in $m/s^2$?

What is retardation?

Define acceleration,average acceleration,and instantaneous acceleration.

The velocity $(v)-$ time $(t)$ plot of the motion of a body is shown below:
The acceleration $(a)-$ time $(t)$ graph that best suits this motion is:

What is acceleration? What is its direction,and state its $SI$ unit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module