The acceleration (a) - time (t) graph for a b | vedclass

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

Question

(A) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.Given that the body starts from rest, the initi

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.Given that the body starts from rest, the initial velocity $(u = 0 \ m/s)$.The area under the $a-t$ graph is a right-angled triangle with base $(b = 10 \ s)$ and height $(h = 8 \ m/s^2)$.Area = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Area = $\frac{1}{2} 	imes 10 \ s 	imes 8 \ m/s^2 = 40 \ m/s$.Since $\Delta v = v_{max} - u = 40 \ m/s$ and $u = 0 \ m/s$, the maximum speed is $v_{max} = 40 \ m/s$.

The acceleration $(a)$ - time $(t)$ graph for a body starting from rest is given below. The maximum speed of the body is (in $m/s$)

Similar Questions

$A$ particle moves along a straight line such that its displacement at any time $t$ is given by $s = (t^3 - 3t^2 + 2) \, m$. The displacement when the acceleration becomes zero is ........ $m$.

$A$ particle starts from rest. Its acceleration $(a)$ versus time $(t)$ graph is as shown in the figure. The maximum speed of the particle will be.....$m/s$.

The acceleration of a particle which moves along the positive $x$-axis varies with its position as shown in the graph. If the velocity of the particle is $0.8 \ m/s$ at $x = 0$,find the velocity of the particle at $x = 1.4 \ m$ (in $m/s$).

Each of the three graphs represents acceleration versus time for an object that already has a positive velocity at time $t_1$. Which graphs show an object whose speed is increasing for the entire time interval between $t_1$ and $t_2$?

In what different ways can velocity be changed?

Vedclass Test Series

Exam Paper Generator

Online Exam Module