Motion Test

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Motion Test - 72

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Weekly Quiz Competition

Question 1
1 / -0

With what speed in miles/hour ($$1 m/s = 2.23 mi/h$$) must an object be thrown to reach a height of $$91.5 m$$ (equivalent to one football field)? Assume negligible air resistance.

A
$$94.4 mi/h$$

B
$$84.4 mi/h$$

C
$$74.4 mi/h$$

D
$$94 mi/h$$

Solution

As we assume negligible air resistance, so the acceleration of the object is due to gravity, i.e. $$a=-g=-9.8 m/s^2$$
where minus sign for motion against gravity.
When the object reaches top most height, the final velocity will be zero i.e $$v=0 \ m/s$$
Maximum height reached $$S = 91.5 \ m$$
Let $$u$$ be the initial velocity of the object.
Using formula $$v^2-u^2=2aS$$
$$\therefore$$ $$0^2-u^2=2(-9.8)(91.5)$$
$$\implies$$ $$u=42.35 m/s=42.35\times 2.23 mi/h=94.4 mi /h$$
Question 2
1 / -0

A race car accelerates uniformly from $$18.5 m/s$$ to $$46.1 m/s$$ in $$2.47$$ seconds. Determine the distance traveled by the car. (in m)

A
$$79.8 m$$

B
$$79 m$$

C
$$7.8 m$$

D
$$9.8 m$$

Solution

Here, initial velocity is $$u=18.5 m/s$$ and final velocity is $$v=46.1 m/s$$
Time taken $$t=2.47 s$$
If $$a$$ be the acceleration of the car.
Using $$v=u+at$$,
$$46.1=18.5+a(2.47)$$
$$\implies$$ $$a=11.17 m/s^2$$
If $$S$$ be the distance traveled by car.
Using formula $$v^2-u^2=2aS$$,
$$(46.1)^2-(18.5)^2=2(11.17)S$$
$$\implies$$ $$S=79.8m$$
Question 3
1 / -0

An engineer is designing the runway for an airport of the planes that will use the airport, the lowest acceleration rate is likely to be $$3 m/s^2$$. The takeoff speed for this plane will be $$65 m/s$$. Assuming this minimum acceleration, what is the minimum allowed length for the runway? (in m)

A
$$704$$

B
$$70$$

C
$$705$$

D
$$707$$

Solution

Given, initial velocity, $$u=0 m/s $$ and final velocity $$v=65 m/s$$
Acceleration $$a=3 m/s^2$$
If $$d $$ be the allowed length for the runway.
Using formula $$v^2-u^2=2aS$$
$$(65)^2-0^2=2(3)d$$
$$\implies$$ $$d=65^2/6=704.16 \sim 704 m$$
Question 4
1 / -0

The driver of a train travelling at $$115\ km\ h^{-1}$$ seen on the same track, $$100\ m$$ in front of him, a slow train travelling in the same direction at $$25\ km\ h^{-1}$$. The least retardation that must be applied to faster train to avoid a collision is

A
$$25\ ms^{-2}$$

B
$$50\ ms^{-2}$$

C
$$75\ ms^{-2}$$

D
$$3.125\ ms^{-2}$$

Solution

Relative velocity of faster train w.r.t. slower train $$u =115 - 25= 90\ km\ h^{-1}$$
$$u= 90\times \dfrac {5}{18} ms^{-1} = 25\ ms^{-1}$$
Using $$v^2 - u^2 =2aS$$
$$0^{2} - 25^{2} = 2\times a\times 100$$
or $$200a = -625 \Rightarrow a = -\dfrac {625}{200} ms^{-2}$$
or $$a = -3.125\ ms^{-2}$$.
Question 5
1 / -0

A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be 10 m/s, then the maximum height attained by the stone is (Take g =10 $$m/s^2$$)

A
3 m

B
15 m

C
1 m

D
10 m

Solution

Let $$u$$ be the initial velocity and $$h$$ be the maximum height attained by the stone.
$$v^2_1 = u^2 - 2gh$$,
$$(10)^2 = u^2 - 2\times 10 \times \frac {h}{2}$$
$$100=u^2 - 10h$$ ....(i)
Again at height $$h$$,
$$v^2_2 = u^2 - 2gh$$
$$(0)^2=u^2-2 \times 10\times h$$
$$u^2 = 20\, h$$ ... (ii)
So, from Eqs. (i) and (ii) we have
$$100 = 10h$$
$$h=10\, m$$
Question 6
1 / -0

The two ends of a train moving with a constant acceleration pass a certain pole with velocities $$u$$ and $$v$$. The velocity with which the middle point of the train passes the same pole is

A
$$\sqrt{\displaystyle\frac{{u}^{2} + {v}^{2}}{2}}$$

B
$$\displaystyle\frac{u + v}{2}$$

C
$$\sqrt{uv}$$

D
$$\displaystyle\frac{{u}^{2} + {v}^{2}}{2}$$

Solution

Let the length of the train $$= l$$ and uniform acceleration $$= f$$

$$\therefore {v}^{2} = {u}^{2} + 2fl$$

$$\Rightarrow f = \displaystyle\frac{{v}^{2} - {u}^{2}}{2l}$$

Let the desired speed be $${v}^{\prime}$$. The train will have moved $$\displaystyle\frac{l}{2}$$ till it attains a speed $${v}^{\prime}$$.

$${v}^{\prime 2} = {u}^{2} + 2f \cdot \displaystyle\frac{l}{2} = {u}^{2} + fl$$

$$ = {u}^{2} + \displaystyle\frac{{v}^{2} - {u}^{2}}{2l} \cdot l$$

$$ = {u}^{2} + \displaystyle\frac{{v}^{2} - {u}^{2}}{2}$$

$$ = \displaystyle\frac{{u}^{2} + {v}^{2}}{2}$$

$${v}^{\prime} = \sqrt{\displaystyle\frac{{u}^{2} + {v}^{2}}{2l}}$$
Question 7
1 / -0

The graph below shows the distance travelled and the time taken by four cars.
Which car travelled the slowest?

A
Car $$1$$

B
Car $$2$$

C
Car $$3$$

D
Car $$4$$

Solution

The slope of the distance-time graph represents the velocity.
The slope for the car $$4$$ is minimum, so its speed is also minimum.
Question 8
1 / -0

A car, moving at 1.5 $$m{s}^{-1}$$ applies brakes and comes to rest in $$2 s$$. If the same car travels at double the speed, what time would it take to come to rest after applying brakes?

A
$$4s$$

B
$$8s$$

C
$$2s$$

D
$$3s$$

Solution

Given,
Initial velocity $$u_1=1.5ms^{-1}$$
Final velocity $$v_1=0$$
Time taken $$t_1=2s$$

According to first equation of motion,
$$v_1=u_1+a_1t_1$$
$$0=1.5+2a_1$$
$$a_1=\dfrac{-1.5}{2} = -0.75\ ms^{-2}$$

Now, for the second case,
Initial velocity $$u_2=3ms^{-1}$$
Final velocity $$v_2=0$$
The braking system is still the same. So it provides the same retarding acceleration of $$-0.75 \ ms^{-1}$$
So acceleration $$a_2$$ is same as $$a_1=-0.75 \ ms^{-1}$$
Using first equation of motion again,
$$v_2=u_2+a_1t_2$$
$$0=3-0.75\times t_2$$
$$t=\dfrac{3}{0.75}=4s$$
Question 9
1 / -0

The figure shows the displacement time graph of a particle moving on a straight line path. What is the magnitude of average velocity of the particle over 10 s?

A
$$2\;ms^{-1}$$

B
$$4\;ms^{-1}$$

C
$$6\;ms^{-1}$$

D
$$8\;ms^{-1}$$

Solution

Net displacement of the particle in 10 s, $$S = BC - OA = 40 -60 = -20$$ m
Total time taken $$t = 10$$ s
$$\therefore$$ Average velocity of the particle $$v_{avg} = \dfrac{S}{t} = \dfrac{-20}{10} = -2 m/s$$
Thus magnitude of average velocity of the particle is $$2 m/s$$.
Question 10
1 / -0

Two cyclists, k km apart, and starting at the same time, would be together In r hours if they travelled in the same direction, but would pass each other In t hours if they travelled in opposite directions. The ratio of the speed of the faster cyclist to that of the slower is ____

A
$$\frac{r+t}{r-t}$$

B
$$\frac{r}{r-t}$$

C
$$\frac{r+t}{r}$$

D
$$\frac{r}{t}$$

Solution

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Re-Attempt Weekly Quiz Competition

Motion Test - 72

Result

Motion Test - 72

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SHARING IS CARING

Question 1

$$94.4 mi/h$$

$$84.4 mi/h$$

$$74.4 mi/h$$

$$94 mi/h$$

Solution

Question 2

$$79.8 m$$

$$79 m$$

$$7.8 m$$

$$9.8 m$$

Solution

Question 3

$$704$$

$$70$$

$$705$$

$$707$$

Solution

Question 4

$$25\ ms^{-2}$$

$$50\ ms^{-2}$$

$$75\ ms^{-2}$$

$$3.125\ ms^{-2}$$

Solution

Question 5

3 m

15 m

1 m

10 m

Solution

Question 6

$$\sqrt{\displaystyle\frac{{u}^{2} + {v}^{2}}{2}}$$

$$\displaystyle\frac{u + v}{2}$$

$$\sqrt{uv}$$

$$\displaystyle\frac{{u}^{2} + {v}^{2}}{2}$$

Solution

Question 7

Car $$1$$

Car $$2$$

Car $$3$$

Car $$4$$

Solution

Question 8

$$4s$$

$$8s$$

$$2s$$

$$3s$$

Solution

Question 9

$$2\;ms^{-1}$$

$$4\;ms^{-1}$$

$$6\;ms^{-1}$$

$$8\;ms^{-1}$$

Solution

Question 10

$$\frac{r+t}{r-t}$$

$$\frac{r}{r-t}$$

$$\frac{r+t}{r}$$

$$\frac{r}{t}$$

Solution

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