Test: Motion in a Straight Line (May 23) - NEET MCQ

Step 1: Velocity

Velocity v is the first derivative of displacement x with respect to time t:

Step 2: Time when the particle comes to rest

A particle comes to rest when its velocity v=0. Setting v to zero:

2at − 3bt² = 0

 t (2a−3bt) = 0

This gives two solutions:

Thus, the particle comes to rest after time t = 2a / 3b.

Step 3: Initial Velocity and Acceleration

• Initial velocity v₀​ is obtained by substituting t=0 in the velocity equation:

So, the initial velocity is zero.

• Acceleration a_net​ is the derivative of velocity:

At t=0, the acceleration is:

Thus, the initial acceleration is not zero.

Step 4: Net Force

The net force acts based on acceleration. Given the acceleration:

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.
 

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.

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

Let A be the angle of projection and u be the velocity of projection.

 The displacement in y-direction is zero.
0 = (u sin q)t - gt²/2
t = 2u sin q/g
R = (2u²sin q cos q)/g
2u²sin q cos q = gR
Multiply by 2 tan q
4u²sin²q = 2gR tan q
(2u sin q)² = 2gR tan q
g²t²= 2gR tan q
gt²= 2R tan q

<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 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

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