The displacement-time graph of two moving particles A and B respectively make angles of 30ยฐ and 45ยฐ with the x-axis respectively. The ratio of their velocities is
Slope of displacement time graph represents velocity.
VA= tan 30 0
VB= tan 45 0
So, VA/VB=tan 30 0/tan 45 0= 1/โ3
A body is travelling in a straight line with a uniformly increasing speed. Which one of the plot represents the change in distance (s) travelled with time (t)?
\[ s= \frac{1}{2}at^{2}\]
As s is proportional to square of t. s-t graph is a parabola.
A particle is thrown upwards, then correct v-t graph will be
Slope of v-t graph represent acceleration. when object is thrown upward its acceleration is -g for the entire journey. So, slope is -g.
A body starts from rest and moves with uniform acceleration. Which of the following graphs represents it motion?
\[ v = u + at ; as, u = 0 ; v = at\]
As, v is proportional to t ; graph of v-t is a straight line.
If the figure below represents a parabola, identify the physical quantities representing Y and X for constant acceleration
A particle is thrown vertically upwards with a velocity ฮฝ. It returns to the ground in time T. Which of the following graphs correctly represents the motion?
Graph of velocity time graph represents acceleration. During the entire journey acceleration is -g. so slope is negative.
A car moving with a speed of 50 km hโยน can be stopped by brakes after atleast 6m. If the same car is moving at a speed of 100 km hโยน the minimum stopping distance is :
The stopping distance is proportional to the square of the speed:
\[
\frac{s_2}{s_1} = \left( \frac{v_2}{v_1} \right)^2
\]
Substituting the values:
\[
\frac{s_2}{6} = \left( \frac{100}{50} \right)^2 = 2^2 = 4
\]
\[
s_2 = 6 \times 4 = 24 \, \text{m}
\]
Thus, the minimum stopping distance at 100 km/h is 24 meters.
The displacement-time graph of a moving particle is as shown in the figure. The instantaneous velocity of the particle is negative at the point :
Slope of x-t graph represents velocity. Slope is negative at point E so, velocity is negative for E.
Assertion (A): For uniformly accelerated motion along straight line, the position versus time graph is a straight line.
Reason (R): For uniformly accelerated motion the position in equal intervals of time changes by same amount.
Assertion (A) is false.
For uniformly accelerated motion, position \( x \) is related to time \( t \) by \( x = x_0 + v_0 t + \frac{1}{2} a t^2 \). This is a parabolic equation, so the position-time graph is a parabola, not a straight line.
Reason (R) is false. In uniformly accelerated motion, velocity changes by equal amounts in equal time intervals, but position does not. Position changes by increasing amounts (if starting from rest).
Therefore, both the Assertion and the Reason are false.
Assertion (A): In one dimensional motion, area under velocity-time graph gives change in position i.e., displacement.
Reason (R): In one dimensional motion, area under acceleration-time graph gives final velocity.
Area under v-t graph = displacement; Area under a-t graph = change in velocity.
Solution: (A) is true as area under v-t graph is displacement. (R) is false as area under a-t graph gives change in velocity, not final velocity. So, (A) is true and (R) is false.