Motion

Displacement is shortest distance between initial point to final point and its direction will be from A to B. So it will be a straight line from point A to point B.

Sentence can be rephrased as “She moves at constant velocity.” or you can also write it as “She moves at zero acceleration.” as there is no change in its speed or direction resulting in no change in velocity.

Given, distance = 100m and time = 4sec.

Using formula,

If distance = 50m and time = 2 sec, then

= 25 m/s

So speed remains same for both the cases.

Correct statement will be: “The car rounded the curve at a constant speed of 70 km/h.”

Explanation:

Because velocity denotes speed (magnitude) with direction. And, we know that in circular motion, the direction changes, as shown below:

So velocity will not remain constant but speed will remain constant.

First ball will reach the bottom first because all the balls will have same initial speed but path covered by the first ball is the shortest one i.e. straight line between top of the hill and bottom.

So, as distance is shorter and initial speed is same, so first ball will take less time than other two balls.

Car A moves faster than car B because slope of s-t graph denotes speed.

Here slope of A is greater than slope of B so car A is moving with faster speed than car B.

Slope of Distance-time graph denotes speed and change in slope of the graph is acceleration. So it must be constant and positive for this graph as speed is increasing.

Slope of Distance-time graph denotes speed and change in slope of the graph is acceleration. So it must be constant and negative for this graph as speed is decreasing.

Let the total running time be t h.

Distance covered in first h

Distance covered in other h

Total distance

So, Average speed

60 km/h

Let the total distance covered by the car be ‘d’ km.

Time taken to cover first distance, t₁

Time taken to cover second distance, t₂

Total time = t₁ + t₂

Therefore, average speed

= 44.44 km/h

Let the total displacement covered after n sec = S_n

Let the initial velocity be u and acceleration be a.

Displacement covered in n^th sec = S_n – S_n-1

From the equation of motion, we know that

S_n

S_n-1

Displacement covered in n^th sec = S_n – S_n-1

()

Let the initial speed be u m/s and speed after 5s be v₁m/s.

Let acceleration of the particle be a m/s².

v₁ = u + 5a …(using v = u + at) …(1)

Now for first 5s, distance s₁

(2)

For next 3s, v₁ will be its initial velocity

and distance s₂

...(3)

Now s₁-s₂ = 0

Putting value of u in equation (2), we get

…(5)

So (from equation (4))

Given, Distance in 8sec = 10m + 10m = 20m

Distance in 10sec

Distance covered in next 2sec

=

= 8.33 m

Given, initial velocity = 0, acceleration = α and retardation = β

Let the time of accelerated motion be t’ and time of retarded motion be t”, so that t’ + t” = t … (1)

Velocity after time t’, v₁ = 0 + αt’ (using v = u + at) …(2)

Now, v₁ will be initial velocity for retarded motion and

final velocity = 0 and acceleration

0 = v₁- βt” …(using v = u + at)

Or, …(3)

Now, in equation (2)

As, after time t’ , retardation will start so maximum velocity will

be at time t’ which is v₁ .

Given, initial velocity of bus, u = 0 and acceleration, a = 1 m/s².

When man catches the bus, distance between them = 0.

Let at time t sec, man catches the bus first.

Distance covered by man in t sec, d₁ = velocity x time

= 10 m/s x t sec

= 10t m

Distance covered by bus in t sec, d₂ m

As man was 48m behind, so difference between their distance covered = 48m

So, Minimum time to catch the bus will be 8sec.

Given, velocity after 5s = 1.5 m/s

Velocity after 6s = 0 m/s

Let the acceleration be ‘a’ and initial velocity be ‘u’.

For t = 5s,

1.5 = u + 5a …(1) (using v = u + at)

For t = 6s

0 = u + 6a …(2) (using v = u + at)

Subtracting equation (2) from (1), we get

1.5 = -a

Or, a = -1.5 m/s²

Putting value of a in equation (1), we get

u = 9 m/s

So, Distance traversed before stopping = Distance covered in 6s

Body will again move with same acceleration in opposite direction,

So here initial velocity = 0, acceleration a = -1.5 m/s² and

displacement s = -27m (direction is opposite)

putting all these values in we get

-27 =

Velocity at point “O” = velocity after 6s

= 0 + (-1.5)x6 (using v = u + at)

= -9 m/s

Constant acceleration means change in velocity per unit time should be constant. i.e.,

It means body should be changing its velocity with equal amount in equal time interval.

No. When the velocity is constant, average velocity over any time interval will not differ from instantaneous velocity at any instant because as velocity is constant, so it will not change during any time interval. It remains same for every point of time. Therefore, average velocity remains equal to the constant velocity.

Yes. Direction of velocity of an object can be reversed when it’s acceleration is constant.

For example: When we throw a ball upward, it’s acceleration will be due to gravity i.e. 9.8 m/s² in downward direction.

After reaching highest point, its direction will be reversed while its acceleration will remain same i.e. due to gravity.

Given, acceleration = a,
initial velocity = 0
time = t

Velocity after time t, v = …(1)

Now it will become initial velocity.

After acceleration change its sign, let time taken by it to come to rest be t’,

…(2)

Total time = t + t’

Now distance covered in total time

=

=

=

=

=

=…(3)

Now after coming to rest it will again start moving in opposite direction. To reach initial point, it has to cover same distance,

So let time taken to cover that distance be t”. Here initial velocity will be 0. Putting value of distance from equation (3), we get.

Time t₀

=

Given, total distance need to cover, d = 2.7km = 270000 cm

Let distance covered in accelerated motion be d₁ and distance covered in decelerated motion be d₂.

So, d₁ + d₂ = d …(1)

Now for d₁, initial velocity will be 0, and acceleration a = 20 cm/s² and let final velocity be v.

Then,

Now for d₁, initial velocity will be v, and acceleration a’ = 100 cm/s² and velocity will be 0.

Then,

Now putting values of d, d₁ and d₂ in equation (1), we get

And also for d_1, let the time taken to travel be t₁ then

And also for d_2, let the time taken to travel be t₂ then

Now total time to travel will be t₁ + t₂

Given, speed = 10m/s

Distance needed to be covered to cross electric pole = length of the train

= 50 m

Time taken to cross electric pole

Distance needed to be covered to cross bridge

= length of the train + length of the bridge

= 50 + 250 m

= 300 m

Time taken to cross electric pole

Given, speed of each train = 30 km/h and

distance between them = 60km

As they are moving in opposite direction and same speed,

So they will crash at middle of the distance between them, i.e. 30km from each train.

Time taken to travel 30 km by each train == 1h

Speed of the bird = 60 km and time = 1h

So distance traveled by bird in 1h = speed × time

= 60 × 1

= 60 km

As trains will come closer to each other, bird can make any number of trips because distance between trains will approach to 0.

So the time taken by bird will also tends to 0. Therefore, bird can make infinite number of trips.