A particle accelerates from rest at a constan | vedclass

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

Question

(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$. Given,total time $T = t_1 + t_2 = 4\, s$. The velocity-

Vedclass · Accepted Answer

$16\, m$

Vedclass · Answer

(B) Let the time taken for acceleration be $t_1$ and the time taken for deceleration be $t_2$. Given,total time $T = t_1 + t_2 = 4\, s$. The velocity-time graph for this motion is a triangle with base $T = 4\, s$ and height $v_{max} = 8\, m s^{-1}$. The distance travelled is equal to the area under the velocity-time graph. Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$. Area $= \frac{1}{2} 	imes 4\, s 	imes 8\, m s^{-1} = 16\, m$. Thus,the total distance travelled is $16\, m$.

$A$ particle accelerates from rest at a constant rate for some time and attains a constant velocity of $8\, m s^{-1}$. Afterwards,it decelerates at a constant rate and comes to rest. If the total time taken is $4\, s$,the distance travelled is:

Similar Questions

$A$ body is thrown vertically upward with velocity $u$,the greatest height $h$ to which it will rise is,

Can the displacement of a particle be zero when the distance travelled is not zero?

If the displacement $-$ time graph for a particle is parallel to the time axis, what is the velocity of the particle?

The speed-time graph of a body is a straight line parallel to the time axis. The body has:

What can you conclude about the motion of a body depicted by the velocity-time graphs $(i)$,$(ii)$ and $(iii)$ given below?

Vedclass Test Series

Exam Paper Generator

Online Exam Module