Test: Motion in a Straight Line - 2 - Grade 11 MCQ

When a particle travels a semicircle, the distance travelled is equal to the semi circumference of the circle while its displacement is equal to the distance between two diametric ends that is diameter 2r.

Given: Dimensions of Hall = 10 m × 10 m × 10 m
To find: Displacement from one corner to another corner
Hall is in the shape of Cube.
Length of longest diagonal of cube is given by,
Diagonal of cube=√3 × Length of edge
Displacement = √3 × 10
Displacement=10√3
So The answer B is correct

Let distance from A to B be x

Time for travel from A to B= x/20

Time for travel from B to A= x/30

Hence,speed for total distance 2x is:

Total distance covered is 2km, and first 1km is covered with speed of 40kmph thus total time taken is 1/40 hrs similarly as second half is travelled with velocity v time taken would be 1/v hrs. But we know that as average velocity is 48 kmph total time travelled can also be written as 2/48 hrs. Thus we get 
1/40 + 1/v = 2/48
I.e 1/v = 1/24 - 1/40
= (40 - 24) / 40 x 24
Thus we get v = 40 x 24 / 16 
= 60 kmph

At some time instant t, the displacement of the particle from origin = √(at² )²+ (bt²)²
I.e. s = t²√a²+b²
Thus we get v =  2t√a²+b²

After each lap done by the car, displacement = 0 m
And distance covered = 2r = 2 x 3.14 x 100 m
Thus avg. velocity = 0 and avg. speed = 628 / 62.8 = 10m/s

Given that s = ½ gt². 
Thus we get v = ds/dt = gt

For the given particle, v = 3t² - 12t + 3
And its acceleration, a = 6t - 12
Thus we get a = 0
At t = 2 sec,
Hence v at t = 2 sec 
V = 3 x 4 - 12 x 2 + 3
= -9m/s

Let's say that the uniform acceleration is a.
Now distance travelled in 0 - 10 sec, x₁ = ½.a x 100 = 50a
Similarly, distance travelled in 0 - 20 sec,  = ½.a x 400 = 200a
Thus we get distance travelled in 10 - 20 sec, x₂ = 200a - 50a = 150a
And, distance travelled in 0 - 30 sec,  = ½.a x 900 = 450a
Thus we get distance travelled in 20 - 30 sec, x₃ = 450a - 200a = 250a
Hence we get the ratio as 1 : 3 : 5

If we will draw a tangent of the point then ot will make angle x with t. Since only the shortest angle will be taken, thus it will also make 180-x.Since we now know that the slope= tan x, thus the slope at E = tan(180-x) =-tanx, thus the instantaneous velocity at E is negative

In the given question C and F have a positive slope so instantaneous velocity is positive. Since point  D has zero slope so instantaneous velocity is zero but E has negative so instantaneous velocity is negative at E.

The distance traveled is equal to the area under the curve which is equal to
S = ½ x 20 x1 + 20 x 1 +½ x 10 x1 + 10 x 2
= 10 + 20 + 5 + 20 
= 55m

For some point of time say time t, the velocities of A and B would be va  and vb respectively.
Hence we see that for both the graphs we get v_a/t = tan 30 and v_b/t = tan 60
Thus we get the required ratio as tan 30 / tan 60 = ⅓ 

Distance = area under curve

=1/2(2+4)×4 + 1/2(2+4)×2

=18m

On a velocity-time graph the slope of the curve represents the acceleration. For OA the slope is 1, while for AB its 0 and for BC its ½.

From the given velocity graph we get that displacement is w=equal to the area under the curve taken with sign while distance is area under the curve after the modulus is taken at first.
Thus displacement is
( ½ x 20 x 2 + 20 x 3 + ½ x 20 x 1) - (½ x 30 x 1) = 90 - 15 = 75m
And the distance will be
( ½ x 20 x 2 + 20 x 3 + ½ x 20 x 1) + (½ x 30 x 1) = 90 + 15 = 105m

We know position time graph is sine graph and velocity is exact differential of position wrt time i.e. differentiation of sine which is negative of cosine. Hence the velocity time graph will be a negative cosine graph.

Let constant acceleration is a

distance travels in 20 seconds 

s = 0*20+a*20²/2

s = 200a -------------(1)

distance travels in first 10 seconds 

s₁ = 0*10+a*10²/2

s₁ = 50a -----------------(2)

distance travels in next 10 seconds 

s₂ = s-s1

s₂ = 200a-50a = 150a ----------------(3)

s₂/s₁ = 150a/50a = 3

As we know initial velocity is zero hence total distance covered, s = ½ g.t² 
I.e s depends upon the t² 
Thus when distance is decreased to ¼ , time decreases to ½ .
I.e 2 sec.

Speed after 4 sec, v = 10 - 2 x 4 
= 2m/s
Thus distance traveled in fifth second, s = 2 x 1 - ½ x 2 x 1
= 2 - 1
= 1m

The velocity after 2 sec = 2a and velocity after 3 sec = 3a
For some constant acceleration a,
Now distance travelled in third second, s₃ = 2a + ½ a
= 5/2 a
Similarly distance travelled in fourth second, s₄= 3a + ½ a
= 7/2 a
Hence the reqiuired ratio is 5/7

h=1/2g(t1_+t2)²

2h/g=(t₁+t₂)²------(1)

 h/2=1/2gt₁²

t₁²=h/g---------(2)

 From 1st and 2nd

t₁²=(t₁+t₂)²/2

t₁=(t₁+t₂)/1.41

1.41t₁=t₁+t₂

t₁=t₂/.41

So

t₁=2.414t₂

Time taken by anybody to fall from a height h is
T = 2gh which is independent of the mass
Hence the required ratio will be that of option c

Acceleration due to gravity takes place every moment on earth whether you are throwing yourself up or else just diving into the pool. It is also indeed true that acceleration causes a change in velocity. But it happens that the change from a negative vector of velocity to positive one undergoes a period when the object/body literally has a Zero velocity for a very very short amount of time.

Total distance =50+50=100
 Total speed=10+15=25 s.
As we know, s= d/t [s = speed (meters/second)
d = distance traveled (meters)
t = time (seconds)] 
we get t= d÷s
so, t=100/25
t= 4 sec

Both the cars are moving in perpendicular directions and thus we stop b and see speed of a wrt b, we get the direction of a is 80.cos 45 + 60.cos 45 east and 80.sin 45 - 60.sin 45.
That is the directions are 140.cos 45 east and 20.sin 45 north.
Thus the angle with north is tan^-1 20/140
= tan^-1 1/7 

v_A​=10m/s
v_B​=5m/s
Here, V_relative=10m/s-5m/s
     =5m/s
Now,t=S/V=100m/5m/s=20sec

Only the horizontal component remains constant as as there is no component of acc. in horizontal direction. The g aaceleration due to gravity acts down words hence there is no accln in horizontal direction, the velocity changes in vertical direction as goes up and down while in horizontal it goes only forward.

We know that horizontal range R depends upon sin 2a where a is angle of projection. Also we know that sin 2a repeats its value for all π/2 -a. thus as here a = 15, the value of range will be equal to the same for 90 -15 = 75.

In the projectile motion the horizontal range of the object is determined as
R = u².sin2x / g
Also we know that range is maximum when angle is 45°
Hence we get R = 22m =  u².sin 90° / g
We get  u² / g = 22
Now when the angle is 15°
We get R = u².sin 30° / g
= u2/ 2g
= 11m