Question

BRAINLIEST !! BRAINLIEST!!​

Answer 1

Answer:

a. Bus accelerated from 0min to 10min

b. Bus decelerated from 10min to 20min

c. Distance covered will be area of triangle from 0min to 10min =

$\frac{1}{2} \times 10 \times 40 = 200m$

d. Distance covered during deceleration is same as that during acceleration since it happens in equal times

= 200 metres.

e. average speed = Distance / Time

$= \frac{200 + 200}{20}$

$= \frac{400}{20}$

$= 20m {</strong><strong>(</strong><strong>min</strong><strong>)</strong><strong>}^{ - 1}$

Answer 2

Answer:

a) 0-10 min

b) 10-20 min

c) distance= v- u / t2 - t1

 = 40-0/10-0

= 40/10=4km/min

d) distance=v-u/t2-t1

= 20-40/10-20

= - 20/ - 10 = 2 km/min

e) average speed=distance/time

                              = 400/20

                              = 20 km/h

I guess this is ur answer...please mark me as the brainliest...