Question

Class-9
physics
Answer fast plese
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Answer 1

Answer:

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Explanation:

1.) We know that,

Speed=Distance / time

.°. Time =dist / speed

Let's see the first case,

s = distance = 30 km

v = speed = 40 km/hr

t = time = 30 / 40

=3/4 hrs

Let's see the second case,

s = 30 km

v = 20 km/hr

t = 30 / 20

t = 3/2 hrs

Avg. Speed =Total distance/ Total time

=(30 + 30) / (3/4) + (3/2)

=60 / ( 3 + 6 ) /4

=60 / 9 × 4

=20 / 3 × 4

=80 / 3

=26.67 ( Approx. )

Hence,

Average speed of the car is 26.67 km/hrs ( Approx. )

2.) (i) In the first case, the train travels at a speed of 60 km/h for a time of 0.52 h

Now, Speed = distance / time = 30km/hr

Distance = 60 x 0.52 = 31.2 km --- i)

(ii) In the second case, the train travels at a speed of 30 km/h for a time of 0.24 hr.

Now, speed = distance / time = 30 km/hr

Distance = 30 × 0.24 = 7.2 km --- ii)

(iii) In the third case, the train travels at a speed of 70km/h for a time of 0.71 hr.

Now, speed = distance / time = 70km/hr

Distance = 70 × 0.71 hr .

Now speed = distance / time = 70km/hr

Distance = 70 × 0.71 = 49.7 km ---iii)

From equation 1,2 & 3 we get,

Total distance travelled

= (31.2 + 7.2 + 49.7) km = 88.1 km .

Total time taken = (0.52 + 0.24 + 0.71) hr = 1.47 hr .

Average speed = Total distance travelled / Total time taken

= (88.1/1.47) km /hr

Hence, average speed = 59.9 km/hr

3.) avg speed = (speed1+speed2)/2

speed1= 250/5

= 50 m/s

speed2= 250/5

= 50m/s

avg speed=( 50+50)/2

= 50 m/s

avg velocity =0 because the displacement is 0 as the bus came back to Delhi covering the same distance

4.) Using first equation of motion

v=u+at, we have

0=20+(−5)t

⇒5t=20

⇒t=4s