Test: Motion in a Straight Line - 3 - Grade 11 MCQ

Particle will return to its starting point when x = 0 by solving we get t = 0 or t = a/b
The particle will come to rest when v = 0 i.e. dx/dt = 0 which is at t = 0 or t = 2a/3b
The initial velocity of the particle = 0 but initial acceleration dv/dt (at t = 0) = 2a
F = ma and a = 2a – 6bt
Acceleration at a/3b = 0 and if a = 0 then F = 0.

If the velocity is zero at an instant, then acceleration may not necessarily be zero at that instant. For e.g. when an object is thrown vertically upward, at highest point velocity is zero but acceleration on it due to gravity is non-zero.

This represents a case that is similar to that of a raindrop attaining terminal velocity.
Fnet = (k1 – k2v)
The acceleration wil become 0 and the velocity will be constant eventually.
Hence B and D

 Time for forward stroke + T_f
Time for return stroke = T_r
R = T_r/T_f
Therefore, time for only one cutting stroke(T) = 1/N x T_f/(T_f + T_r)
Average cutting speed = S/T = SN(T_f + T_r)/T_f = SN(1 + R).

v=change in displacement/change in time ​ and
a=change in velocity/change in time ​
Speed is the magnitude of velocity.
There can be a case where there is change in direction but not in magnitude.

A can be zero when velocity and vice-versa you can get that when you see a pendulum at mean a = 0 but v = max and at extreme a = max when v = 0. Also a “can” be zero when v is zero.

When the ball is at its peak possible position, it's velocity remains Zero. Unfortunately, this moment of zero velocity is hardly visible, since it occurs in a very very short amount of time.
Now if you ask about the acceleration at the peak possible position, it turns out to be very interesting. The acceleration remains 9.81 ms^-2. In fact it is one of the very rare situations where the body is accelerated, still there is an absolute zero velocity.
 

Option A: Acceleration varies as slope of this graph; circular motion does not have the acceleration shown by this graph.
Option B: Acceleration is zero at t=1s, when slope is 0.

1. As it crossed the x axis so the direction changed
2. The slope of v-t graph gives acceleration so it is constant
3. The net areas below and above x-axis are equal so the displacement is zero
4. Speed is a scalar quantity hence the initial and final speeds are same

ATQ: u = 2 m/s, v = -4 m/s, so,
Now, to find the time at which the displacement is zero


Now, to find the displacement at time t = 4 secs wrt the ground

And the displacement at time t = 4 secs wrt the belt

Given constant speed this question so speed is same, since it moved on a straight line so, velocity is same and in the same direction.

The ball will return to him if the cart moves with a constant velocity because the ball will travel the same amount of distance as the cart in between the time interval when it moves up and returns to the same horizontal level. But it will fall behind if the cart has acceleration because then the amount of distance moved by the cart will be more.

R=u2sin2θ=√3
(√20)2sin2θ/10=√3
sin2θ=√3/2
2θ=60o,120o
(A)θ=30o
μ=u²sin²θ/2g=(√20)2sin²30/2x10=0.25
θ=60°
μ=(√20)2sin²30/2x10=3x0.25
(B) U_min=ucosθ=√20cos30°=√5m/s
(C) T=2usinθ/g  For θ=30°
��=2x√20sin²60°/10=√3/5
(D)U_max=mgh=1x10x0.25=2.5J

Let A be the angle of projection and u be the velocity of projection.
(A) Maximum height = u²/2g
Range = (u²sin 2A)/g
Maximum range (at A = π/4) = u²/g
So, max range = 2 × max height

(B) Max Height = u²sin²A/2g
Range = (2u²sin A cos A)/g
Range = n × Max Height
(2u²sin A cos A)/g = n × (u²sin²A)/2g
4sin A cos A = n × sin²A
tan A = 4/n

(C) The displacement in y-direction is zero.
0 = (u sin A)t - gt²/2
t = 2u sin A/g
R = (2u²sin A cos A)/g
2u²sin A cos A = gR
Multiply by 2 tan A
4u²sin²A = 2gR tan A
(2u sin A)² = 2gR tan A
g²t²= 2gR tan A
gt²= 2R tan A

(D) For vertical motion, t = 2u/g
For projectile of, T = 2u sin A/g
If T = t it means sin A = 1 => A = π/2
So, both are vertical motions and hence thier maximum heights will be same (or) their ratio is 1:1

Correct Answer :- b

Explanation : Since the x−t graph is parabola which is of second degree the v−t graph must be a straight line which is of first degree.

The slope of the graph in the period 0≤t<T is positive and decreasing which implies that the velocity is also positive and decreasing in this period.

The slope of the graph at t=T is 0 which implies that the velocity is also 0 at this time.

The slope of the graph in the period T≤t≤2T is negative and decreasing which implies that the velocity is also negative and decreasing in this period.

The only graph which satisfies the above conditions is graph b.

<sub>The acceleration of the particle moving in straight line is given as, a=d2x​/dt2
Geometrically it is the concavity of the x−t graph. If the graph is concave upward then the double derivative is positive. If the graph is concave downwards then the double derivative is negative.Consider the provided x−t graph,
Since the x−t graph is a parabola which of second degree the a−t graph must be a constant line which is of zero degree. Since the parabola is concave downwards the acceleration must be negative.
The only graph which satisfies the above conditions is graph D.

Hence, option D is correct.</sub>

The speed of the particle moving in straight line is always positive and given as, s=∣dx/dt​∣

Geometrically it is the modulus of the slope of the x−t graph.

Consider the provided x−t graph,

Since the x−t graph is a parabola which is of second degree the s−t graph must be a straight line which is of first degree.

The slope of the graph in the period 0≤t<T is positive and decreasing which implies that the speed is positive and decreasing in this period.

The slope of the graph at t=T is 0 which implies that the speed is also 0 at this time.

The slope of the graph in the period T≤t<2T is negative and decreasing which implies that the speed is positive and decreasing in this period.

The only graph which satisfies the above conditions is graph C.

The particle comes to rest (v = 0) at two instants. It comes to rest first time at somewhere between 4th second and 6th second. At t = 8 second it again comes to rest.
We will find when the particle comes to rest for the first time. Between 4th and 6th second (time interval of 2 seconds) the particle changes it's velocity from 10 m/s to -20 m/s. So, the acceleration on the particle during this part of motion is,
a = (-20 - 10)/2 = -15 seconds
The particle comes to rest when v = 0
v - u = at
0 - 10 = (-15)t
⇒ t = 1.5 seconds
So the particle come to rest at (4 + 1.5)th = 5.5 th second and 8th second

The speed is traveled distance (60 miles) divided by traveled time (4pm – 2pm = 2hours):
60 miles/ 2 hours = 30 mph

Displacement of the particle in 2 seconds is the area under v-t graph till 2 seconds,
s = (1/2)(10)(2) = 10 m
Net displacement = -15 + 10 = -5 m

The area under the v-t graph gives displacement. The maximum displacement is when the particle velocity becomes zero and it changes its direction. Since we don't know when the particles crosses x-axis (v = 0) we will first find displacement till 4 seconds
S = (1/2)(10)(2) + (10)(2) = 30 m
The particle velocity changes from 10 m/s to - 20 m/s in next 2 seconds. So it's acceleration during this part of motion is,
a = (-20 - 10)/2 = - 15 m/s²
Distance travelled till velocity becomes zero can be calculated by using the formula v² - u² = 2as
0 - (10)² = 2(-15)s
=> s = 100/30 = 3.33 m
Therefore the maximum displacement is,
d = 30 + 3.3 = 33.3 m
 

Total distance is the entire area under the curve which is area of trapezium + area of triangle.
Trapezium = |½ x 10 x (5 + 2)| = 35
Triangle = |½ x -20 x 2| = 30
Distance = 65m

Displacement is said to the shortest path covered by a body or object. Hence, when the velocity is positive the object moves in one direction and when its negative then in the opposite direction thus making the displacement first increase then decrease.