"The direction in which an object moves is given by the direction of velocity of the object and not by the direction of acceleration." Give an example to justify this statement.
(N/A) When an object is thrown vertically upwards, its velocity is directed upwards because it is moving away from the ground. However, the acceleration due to gravity $(g)$ acts downwards towards the center of the Earth. Since the object continues to move upwards despite the downward acceleration, it proves that the direction of motion is determined by the velocity and not by the acceleration.
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Similar Questions
The slope of the line on a position-time graph reveals information about an object's velocity. What conclusion can you draw regarding the motion of an object,if the graph is a
$(i)$ Horizontal line.
$(ii)$ Straight diagonal line.
$(iii)$ Curved line.
$(i)$ Horizontal line.
$(ii)$ Straight diagonal line.
$(iii)$ Curved line.
Medium
View Solution$(a)$ $A$ car moving with uniform velocity $u$ and uniform acceleration $a$ covers a distance $S$ in time $t$. Draw its velocity-time graph and derive an expression relating all the given physical quantities.
$(b)$ $A$ boy revolves a stone tied to a string $0.7 \, m$ long. Find the distance and displacement covered by the stone in completing two revolutions from the starting point.
$(b)$ $A$ boy revolves a stone tied to a string $0.7 \, m$ long. Find the distance and displacement covered by the stone in completing two revolutions from the starting point.
Difficult
View Solution$A$ body can have zero average velocity but not zero average speed. Justify.
Medium
View SolutionAn object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time.
Time $(s)$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ Velocity $(m s^{-1})$ $2$ $4$ $6$ $8$ $10$ $12$ $14$
Plot the graph.
From the graph:
$(i)$ Find the velocity of the object at the end of $2.5 \ s$.
$(ii)$ Calculate the acceleration.
$(iii)$ Calculate the distance covered in the last $4 \ s$.
| Time $(s)$ | $0$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ |
| Velocity $(m s^{-1})$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ |
Plot the graph.
From the graph:
$(i)$ Find the velocity of the object at the end of $2.5 \ s$.
$(ii)$ Calculate the acceleration.
$(iii)$ Calculate the distance covered in the last $4 \ s$.
Medium
View SolutionName the physical quantities denoted by:
$(i)$ The slope of the distance$-$time graph.
$(ii)$ The area under a velocity$-$time graph.
$(iii)$ The slope of a velocity$-$time graph.
$(i)$ The slope of the distance$-$time graph.
$(ii)$ The area under a velocity$-$time graph.
$(iii)$ The slope of a velocity$-$time graph.
Easy
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