The motion of a lift (elevator) in a shopping centre is given by the velocity-time graph below, where time  `t`  is in seconds, and the velocity of the lift is `v` metres per second. For  `v > 0`  the lift is moving upwards.
 

The graph shows that at the end of 30 seconds, the position of the lift is

A.   17.5 metres above its starting level.

B.   5 metres above its starting level.

C.   at the same position as its starting level.

D.   5 metres below its starting level.

E.   17.5 metres below its starting level.

The velocity–time graph for the first 2 seconds of the motion of a particle that is moving in a straight line with respect to a fixed point is shown below.

The particle’s velocity `v` is measured in cm/s. Initially the particle is  `x_0` cm from the fixed point.

The distance travelled by the particle in the first 2 seconds of its motion is given by

A.   `int_0^2 v\ dt`

B.   `int_0^2 v\ dt + x_0`

C.   `int_1^2 v\ dt - int_0^1 v\ dt`

D.   `|\ int_0^2 v\ dt\ |`

E.   `int_1^2 v\ dt - int_0^1 v\ dt + x_0`

The velocity–time graph below shows the motion of a body travelling in a straight line, where `v\ text(ms)^(−1)` is its velocity after `t` seconds.

The velocity of the body over the time interval  `t in [4,9]`  is given by  `v(t) = −9/16(t - 4)^2 + 9`.

The distance, in metres, travelled by the body over nine seconds is closest to

A.   45.6

B.   47.5

C.   48.6

D.   51.0

E.   53.4

The velocity–time graph for a body moving along a straight line is shown below.
 

The body first returns to its initial position within the time interval

A.   (0, 0.5)

B.   (0.5, 1.5)

C.   (1.5, 2.5)

D.   (2.5, 3.5)

E.   (3.5, 5)