Question

Please help me with the answer​

Answer 1

1) the distance travelled by the object in the first two seconds is equal to the area of ∆ BAE

distance

$= \frac{1 \times base \times height}{2}$

$= \frac{2 \times 15}{2} \\ = 15m$

2)BC

3)

the distance travelled by the object during retardation is equal to the area of ∆ CFD

distance

=

$= \frac{1 \times 15}{2} \\ = 7.5m$

I hope this helps ( ╹▽╹ )

Answer 2

Answer:

(a) . distance traveled during first 2 second

= area of triangle till 2 seconds.

= 1/2 * 2 *15

= 15m.

(b) clearly visible that object has a constant velocity from 2 second to 5 second. because velocity vs time graph is straight line with slope equal to 0 during this interval of time.

(c) distance traveled during retardation is area of graph from time t= 5 to t= 6 seconds.

• it is triangle, so we use area of triangle formula.
• so distance= 1/2* 15*(6-5)

= 15/2

= 7.5 m