Physics, asked by khushidahiya2007, 8 months ago

questions of motion chapter (other than NCERT)​

Answers

Answered by kmathavan4
1

Answer:

Question 1

An air-plane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance travelled before take off.

Answer

Given, the initial velocity = 0 m /s, acceleration = 3.20 m/s2, t = 32.8 s

This is quite a basic question of 2nd equation of motion s=ut+12at2s=ut+12at2

s = 1/2 at2

s=(.5)(3.20)(32.8)2= 1721.344m

Question 2

A Jeep starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the Jeep.

Answer

Use 2nd equation of motion

s=ut+12at2s=ut+12at2

110 = [0.5a × (5.21)22]

a = 110 / (5.2122 × 0.5)

a = 8.10m/s2

Question 3

John is riding the Giant Drop at Canada. If John free falls for 2.6 seconds, what will be his final velocity and how far will he fall?

Answer

Using 1st equation

v=u+at

v=gt where g =acceleration due to gravity=9.8 m/s2

v=9.8X 2.6=25.48 m/s

Using 2nd equation

s=ut+12at2s=ut+12at2

s=12gt2s=12gt2

s=33.124m

Question 4

A racing car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance travelled.

Answer

a=(v−u)ta=(v−u)t

a=(46.1−18.5)2.47a=(46.1−18.5)2.47

a=11.47m/s2

 Question 5

A feather is dropped on the planet other than earth which has very low acceleration due to gravity from a height of 1.40 meters. The acceleration of gravity on the other planet is 1.67 m/s2. Determine the time of feather to fall to the surface of the other planet

Answer

Here u=0, a=1.67m/s2 , s=1.40 , t=?

Using 2nd equation

s=ut+12at2s=ut+12at2

t= 1.29 s

Question 6

Rocket-powered sleds are used to test the human response to acceleration. If a rocket- powered sled is accelerated to a speed of 444 m/s in 1.8 seconds, then what is the acceleration and what is the distance that the sled travels?

Answer

Here u=0, v=444m/s ,t=1.8

a=v−ut=4441.8=246.66m/s2a=v−ut=4441.8=246.66m/s2 Now

s=ut+12at2s=ut+12at2

s=399.6 m

Question 7

Honda Activita accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike.

Answer

Here v=7.10 m/s , s=35.4 m ,u=0 ,a=?

Now,

v2=u2+2asv2=u2+2as

a=.712 m/s2

Question 8

An Aeronautics engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The take off speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

Answer

Here v=65 m/s , s=? ,u=0 ,a=3m/s2

Now,

v2=u2+2asv2=u2+2as

distance= 704m

Question 9

A BMW car travelling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration)

Answer

u = 22.4 m/s,v = 0, t = 2.55 s

using First equation of motion

v = u+at

0 = 22.4+aX2.55

a = -8.784 m/s2

Now using third equation of motion

v2=u2+2asv2=u2+2as

0=(22.4)2 -2X8.784Xs

s=28.56m

 Question 10

A kangaroo is capable of jumping to a height of 2.62 m. Determine the take off speed of the kangaroo.

Answer

Here v=0 , s=2.62m ,u=? ,a=-9.8 m/s2

Now,

v2=u2+2asv2=u2+2as

u= 7.17m/s

Question 11

If Rahul has a vertical leap of 1.29 m, then what is his take off speed and his hang time (total time to move upwards to the peak and then return to the ground)?

Answer

Here v=0 , s=1.29m ,u=? ,a=-9.8 m/s2

Now,

v2=u2+2asv2=u2+2as

u= 5.03m/s.

Now time going Up,

v=u+at

or

t=-u/a = 5.03/9.8 =.513 s

Total hang time = 2 * .513 =1.03s

Question 12

A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the barrel of the riffle, the bullet moves a distance of 0.840 m. Determine the acceleration of the bullet (assume a uniform acceleration).

Answer

Here v=521 m/s , s=.840m ,u=0 ,a=-?

Now,

v2=u2+2asv2=u2+2as

a=162×103a=162×103 m/s2

Question 13

A baseball is popped straight up into the air and has a hang- time of 6.25 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the peak is one- half the total hang-time.)

Answer

Time to rise = 6.25/2= 3.125 s s=? ,a =9.8 m/s2 Now,

mark me as brainlist

Answered by priyankajha12
2

surely many questions are there

and I am giving from icse and west Bengal boards

questions which I am giving is not only from motions but the chapter force and motion ok

Attachments:
Similar questions