1. x t x2
At time t1 the object cannot be at two positions x1 and x2 at once. x1 t1

2. For an object in motion in X direction with uniform acceleration, the connection between displacement and time is
If a  0, the graph x-t is a parabola. But there is no guarantee that the curves in figures A and D are parabolas.
If a = 0, then x = u t which a straight line with the slope u corresponding to the constant velocity of the object. E x t

3. The bullet is stopped with a constant acceleration a
The bullet is stopped with a constant acceleration a. We can use the equations of uniform accelerated motion. The stopping distance s is
You can check the dimensions of each equation. Any equation containing the mass M must be incorrect, since, the unit for the displacement is [m].

4. A ball rolling down a ramp has a uniform acceleration, the ball gets faster and faster.
An object moving in a circle with constant speed, is always changing its direction. So, it is accelerating. The direction of the acceleration is directed towards the centre of the circle. Hence, its velocity is not constant.
An object at the end of a pendulum is always changing its speed and direction. The direction of the acceleration is directed towards the centre of its swing.
If and object is travelling in a straight line with a constant speed, it is impossible for its direction to change. So, its velocity must be constant.

5. In each case, after a time of one period, the objects all returns to their original position. So, all displacements are zero. So, their average velocities must also be zero.
average velocity

6. Acceleration is uniform

7. A = (4+2+4+4-4-4-2) m = + 4 m displacement = + 4 m 4 2 v [ m.s-1 ] 2 4
(1/2)(2)(2)=2
displacement = + 4 m 2 v
[ m.s-1 ]
(1/2)(2)(4)=4
(2)(2)=4
(2)(2)=4 2 4 6 8 10
t [ s ]
(2)(-2)=-4
(1/2)(2)(-4)=-4 -2
(1/2)(2)(-2)=-2 -4

8. = (4+2+4+4+|-4|+|-4|+|-2|) m = 24 m
distance travelled
= ( |-4|+|-4|+|-2|) m = 24 m
(1/2)(2)(2)=2 2 v
[ m.s-1 ]
(1/2)(2)(4)=4
(2)(2)=4
(2)(2)=4 2 4 6 8 10
t [ s ]
(2)(-2)=-4
(1/2)(2)(-4)=-4 -2
(1/2)(2)(-2)=-2 -4

9. B A 30 v [ m.s-1 ] (60 - t1) (30) (1/2) (t1) (30) t1 60 t [ s ]
area under curve = displacement
acceleration = slope v-t graph B 30 v
[ m.s-1 ]
(60 - t1) (30)
(1/2) (t1) (30) A t1 60
t [ s ]

10. +X
starting position

12. A B C x t x t x t D x t E x t

14. 4 2 v
[ m.s-1 ] 2 4 6 8 10
t [ s ] -2 -4

15. 4 30 v
[ m.s-1 ] B A
t [ s ] 60