The graph of position x versus time t represe | vedclass

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

Question

(C) The position of the particle is given by the equation $x(t) = a + bt - ct^2$.To find the velocity $v(t)$,we take the first derivative of position

Vedclass · Accepted Answer

$a_{acc} = -2c$

Vedclass · Answer

(C) The position of the particle is given by the equation $x(t) = a + bt - ct^2$.To find the velocity $v(t)$,we take the first derivative of position with respect to time:$v(t) = \frac{dx}{dt} = \frac{d}{dt}(a + bt - ct^2) = b - 2ct$.To find the acceleration $a_{acc}(t)$,we take the derivative of velocity with respect to time:$a_{acc}(t) = \frac{dv}{dt} = \frac{d}{dt}(b - 2ct) = -2c$.Since $c$ is a positive constant,the acceleration is a constant negative value. Thus,the correct expression is $a_{acc} = -2c$.

Similar Questions

$A$ point moves in a straight line so that its displacement $x \ m$ at time $t \ s$ is given by $x^2 = 1 + t^2$. Its acceleration in $m/s^2$ at a time $t \ s$ is

When are the average and instantaneous accelerations equal?

The distance-time graph of a particle at time $t$ makes an angle of $45^{\circ}$ with the time axis. After one second,it makes an angle of $60^{\circ}$ with the time axis. What is the average acceleration of the particle?

What is retardation?

Which physical quantity can be found by first differentiation and second differentiation of position vector?

Vedclass Test Series

Exam Paper Generator

Online Exam Module