Motion Test

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Motion Test - 59

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Question 1
1 / -0

The graphs below show the position versus time for three different cars: $$1, 2,$$ and $$3$$.
Rank these cars according to the magnitudes of their velocities at the time "t" indicated on the graph, greatest first.

A
$$1, 2, 3$$

B
$$1, 3, 2$$

C
$$2, 1, 3$$

D
$$3, 2, 1$$

E
$$3, 1, 2$$

Solution

The velocity in a position-time graph is given by the slope of the graph with the time axis, as the graph of 3 has zero slopes therefore graph 3 shows zero velocity.
Now we can see that slope of graph 2 is greater than graph 1 therefore velocity by graph 2 will be greater than graph 1.
The result is 2,1,3.
Question 2
1 / -0

The graph above shows position versus time for an object moving along a straight line. During which time interval(s) is the object's speed increasing?
I. $$0$$ to $$1$$ seconds
II. $$1$$ to $$2$$ seconds
III. $$2$$ to $$3$$ seconds

A
I only

B
II only

C
III only

D
I and III only

Solution

Slope of the position-time graph gives the instantaneous velocity of the object.
During $$0<t<1$$ s : Slope of the curve is constant, thus the velocity of the object is also constant.
During $$1<t<2$$ s : Slope of the curve is zero, thus the velocity of the object is zero.
During $$2<t<3$$ s : Slope of the curve is positive and increasing, thus the velocity of the object increases during $$1$$ to $$2$$ seconds.
Hence option C is correct.
Question 3
1 / -0

The two position-time graphs time in seconds pictured below on the same grid represents the position of a motorized car (red) and a cart coming down a ramp (blue). At what time are their velocity close to being the same?

A
1 second

B
2 second

C
3 second

D
4 second

E
5 second

Solution

Velocity of an object is given by the slope of its displacement-time or position-time graph,
The slopes of the two curves becomes parallel at approx.at $$t=3s$$.
So the slopes of the two curves is same at $$t=3s$$
Hence the velocities of the two objects is same at $$t=3s$$
Question 4
1 / -0

Which of the following vehicles is undergoing a deacceleration?

A
A car driving straight to the east on a road at a constant speed

B
A truck rounding a corner at a constant speed

C
A van slowing down as it approaches a stop sign

D
None of these

Solution

A object is said to have an acceleration if it changes its velocity either by increasing its speed, decreasing its speed or changing the direction of its velocity.
Since the car and the truck move with constant speed, thus they have zero acceleration.
But the van is slowing down its speed, thus it has deacceleration.
Hence option C is correct.
Question 5
1 / -0

A purple car is moving three times as fast as a yellow car. Each car slows down to a stop with the same constant acceleration. How much more distance is required for the purple car to stop ?

A
twice as much distance

B
three times as much distance

C
five times as much distance

D
seven times as much distance

E
nine times as much distance

Solution

Let the initial velocity of purple car be $$3u$$ , so the initial velocity of yellow car will be $$u$$. Final velocities ( $$v_{p} ,v_{y}$$ )of both cars are zero , acceleration $$a$$ is same for both cars .
Using, the third equation of motion $$v^{2}=u^{2}-2as$$

For purple car $$0=(3u)^{2}-2as_{p}$$
where $$s_{p}=$$ distance required for purple car to stop ,
$$a=$$ acceleration (deceleration)
Hence, $$s_{p}=9u^{2}/2a$$ ,

For yellow car $$0=u^2-2as_y$$
Hence, $$s_{y}=u^{2}/2a$$
where $$s_{y}=$$ distance required for yellow car to stop ,

now by dividing $$s_{p}$$ from $$s_{y}$$
we get $$s_{p}/s_{y}=9$$
or $$s_{p}=9s_{y}$$

Hence, the purple car requires nine times as much distance as the yellow car to stop.
Question 6
1 / -0

Each of the following graphs represents the position of a car as it moves or stays along a straight line. The scales on all the graphs are the same.
Which graph represents the fastest speed of the car at time T?

A

B

C

D

E

Solution

In a position-time graph, the slope of the graph gives the speed of the object at that point. In graphs B, C, and E slope seems zero at T, therefore speed is also zero. Now if we compare graphs A and D, the slope of graph D at T is greater than that in A, therefore speed in graph D is maximum.
Question 7
1 / -0

The velocity graphs below represent the motion of $$5$$ different object moving along a north-south line.
Which object changed directions during the represented movement?

A

B

C

D

E

Solution

As we can see from the given graphs , in graph $$C$$ the velocity is positive as well as negative , when the sign of velocity changes it means change in direction , therefore in graph C , velocity changes direction .
Question 8
1 / -0

The two ends of a train moving with uniform acceleration pass a certain point with velocities $$6\ kmph$$ and $$8\ kmph$$ respectively. What is the velocity with which the middle point of the train passes the same point?

A
$$7\ kmph$$

B
$$5\sqrt{2}\ kmph$$

C
$$2 \sqrt{10}\ kmph$$

D
$$7.5\ kmph$$

Solution

The rear point which had initially a velocity of $$6\ kmph$$ attained a velocity of $$8\ kmph$$ after having a displacement of the length of the train's

Using
$$v^2 –u^2 = 2as $$

$$as = \dfrac{(8^2 -6^2)}{2}=14$$

The middle point which had initially a velocity of $$6\ kmph$$ will acquire a velocity $$V$$ after moving through a distance equal to half the length of the train $$= (s/2)$$

$$V^2 =u^2 + 2 a (s/2)$$
$$\Rightarrow V^2= u^2 + as$$
$$\Rightarrow V^2 = 6^2 + 14 $$
$$\Rightarrow V^2= 50 $$

$$V = 5\sqrt2 kmph $$
Question 9
1 / -0

An airplane accelerates down a runway at $$3.2 m s^{-2}$$ for 32.8 s until it finally lifts off the ground. Determine the distance travelled before takeoff.

A
1323 m

B
1527 m

C
1721 m

D
1931 m

Solution

Given,
$$u=0\ m/s\\a=3.2\ m/s^{-2}\\t=32.8\ s\\ s=?$$
From equation of motion
$$s=ut+\dfrac{1}{2}at^2$$

$$s=0+\dfrac{1}{2} \times 3.2\times 32.8^2=1721.34\ m$$
Question 10
1 / -0

When will a body have zero speed?

A
When a body has uniform acceleration

B
When a body has non-uniform acceleration

C
When a body is at rest

D
When a body is under motion

Solution

The distance travelled by a body per unit time is known as speed.
$$speed=\dfrac{distance}{time}$$
A body has zero speed when it has zero distance covered in an interval of time and hence the body is at rest.