Question

2. How far will a bird have traveled in 13 seconds if it maintained a speed of 31 m/s?

Answer 1

#2

speed of bird = 31 m/s

time = 13 s

since here bird maintains its speed to be constant

so here the distance traveled is given by

$d = v*t$

$d = 31 * 13$

$d = 403 m$

so bird will travel 403 m distance while maintaining its speed to be constant

#3

average speed is the ratio of total distance covered and total time

here Ian walk 2 km then 0.5 km and then 2.5 km so the total distance is given as

$d = d_1 + d_2 + d_3$

$d = 2 + 0.5 + 2.5$

$d = 5 km$

total time of his full motion is

$t = 3 hours$

so the average speed of the motion is

$v = \frac{d}{t}$

$v = \frac{5}{3} = 1.67 km/h$

so here average speed of Ian is 1.67 km/h

#4

Initial speed of the truck is 6 m/s

after 4 s interval of time

final speed of the truck is 14 m/s

so the acceleration is rate of change in velocity

so it is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{14 - 6}{4}$

$a = 2 m/s^2$

so here truck accelerated with 2 m/s^2

#5

Here car starts from rest so its initial speed is 0

after 2.5 s interval of time

final speed of the car is 15 m/s

so the acceleration is rate of change in velocity

so it is given as

$a = \frac{v_f - v_i}{t}$

$a = \frac{15 - 0}{2.5}$

$a = 6 m/s^2$

so here car accelerated with 6 m/s^2

#6

a bag full of trash is dropped from the balcony

Here we can say when bag is in air

the net force on the bag is only due to its weight

so here we can use Newton's II law to find the acceleration of bag

$F = ma$

$mg = ma$

$a = g = 9.8 m/s^2$

so here bag will have constant acceleration during its whole motion as we can see that there is no change in the force and hence its acceleration always remains constant