Unit Test (Solutions): Motion

Time: 1 hour 
M.M.: 30 
Attempt all questions. 
Question numbers 1 to 5 carry 1 mark each. 
Question numbers 6 to 8 carry 2 marks each. 
Question numbers 9 to 11 carry 3 marks each. 
Question numbers 12 & 13 carry 5 marks each.

Q1: An object travels 16 meters in 4 seconds and then another 16 meters in 2 seconds. What is the average speed of the object? (1 Mark)
(i) 4 m/s
(ii) 5.33 m/s
(iii) 6 m/s
(iv) 8 m/s

Ans: (ii) 
To find the average speed, we use the formula:
Average Speed=Total Distance/ Total Time
The object covers a total distance of 16 m + 16 m = 32 m.
The total time taken is $4s+2s=$6 s.
Thus, the average speed is:

$Average Speed=\frac{32}{6}\approx 5.33m/s\backslash text\{Average\; Speed\}\; =\; \backslash frac\{32\; \backslash ,\; \backslash text\{m\}\}\{6\; \backslash ,\; \backslash text\{s\}\}\; \backslash approx\; 5.33\; \backslash ,\; \backslash text\{m/s\}$ m/s

Q2: What does the slope of a distance-time graph represent? (1 Mark)
(i) Speed
(ii) Acceleration
(iii) Displacement
(iv) Velocity

Ans: (i)

Q3: If a car travels 50 km in the first hour and 70 km in the next two hours, what is its average speed? (1 Mark)
(i) 40 km/h
(ii) 60 km/h
(iii) 50 km/h
(iv) 30 km/h

Ans: (i) 
Average speed: Total Distance/ Total Time

Total distance covered = $50km+70km=\mathrm{120 }km50\; \backslash ,\; \backslash text\{km\}\; +\; 70\; \backslash ,\; \backslash text\{km\}\; =\; 120\; \backslash ,\; \backslash text\{km\}$km.
Total time taken = $1hour+2hours=$3 hours.
Thus, the average speed is:

$Average Speed=\frac{120}{3}=40km/h\backslash text\{Average\; Speed\}\; =\; \backslash frac\{120\; \backslash ,\; \backslash text\{km\}\}\{3\; \backslash ,\; \backslash text\{hours\}\}\; =\; 40\; \backslash ,\; \backslash text\{km/h$ km/h

Q4: Which of the following graphs represents an object at rest? (1 Mark)

Ans: A
A straight line parallel to the x-axis on a distance-time graph represents that the object is at rest.

Q5: The area under a velocity-time graph represents which of the following quantities? (1 Mark)
(i) Speed
(ii) Acceleration
(iii) Distance travelled
(iv) Time taken

Ans: (iii) 
The area under a velocity-time graph represents the distance travelled by the object. This is because multiplying velocity (y-axis) by time (x-axis) gives the displacement (distance if direction is not considered). For constant velocity, this area is a rectangle, while for changing velocity, it could be a combination of rectangles, triangles, or other shapes. 

Q6: A car accelerates from rest at a constant rate of 3 m/s² for 8 seconds. What is the final velocity and the distance travelled by the car? (2 Marks)
Ans: As the car initially is at rest, so u = 0 m/s

Final velocity (v) = $u+at=$0+(3m/s²×8s) = 24m/s

Distance travelled (s) = $ut+\frac{1}{2}at\mathrm{<sup\; style="font-size:12px\; !important">2</sup>}=0+\frac{1}{2}\times 3m/s\mathrm{<br><br>}\times \left(8s\right)\mathrm{<sup\; style="font-size:12px\; !important">2</sup>}=$96m

Q7: Differentiate between Distance and Displacement motion using examples. (2 Marks)
Ans: 

Q8: Explain the concept of uniform circular motion. Derive the formula for the speed of an object in circular motion and give examples. (2 Marks)
Ans:

Concept: In uniform circular motion, an object moves with constant speed along a circular path. The velocity is always tangential to the circle, and the direction changes continuously, implying constant acceleration towards the centre.
Formula Derivation:
The circumference of the circle = $2\pi r2\backslash pi\; r$
Speed $v=\frac{2}{\pi }$, where $T$ is the time period.
Example: The motion of a satellite orbiting the Earth, the motion of a car on a circular track.

Q9: Write the three equations of motion and explain their significance with examples. (3 Marks)
Ans:

First equation of motion: $v=u+at$
It relates the final velocity to initial velocity, acceleration, and time. 
Example: A car accelerating from rest.
Second equation of motion: $s=ut+\frac{1}{2}at\mathrm{<sup\; style="font-size:12px\; !important">2</sup>}$
It gives the displacement of an object in terms of time, initial velocity, and acceleration. 
Example: Distance covered by a freely falling object.
Third equation of motion: $v\mathrm{<sup\; style="font-size:12px\; !important">2</sup>}=u\mathrm{<sup\; style="font-size:12px\; !important">2</sup>}+2asv^2\; =\; u^2\; +\; 2as$
It relates the final velocity to initial velocity, acceleration, and displacement. 
Example: Calculating the speed of a car after braking.

Q10: Plot a distance-time graph for an object moving with uniform speed and another for an object moving with non-uniform speed. Explain the difference. (3 Marks)
Ans: 

For uniform speed: The distance-time graph will be a straight line inclined with the time axis, showing that distance increases uniformly with time.

For non-uniform speed: The graph will be a curve indicating that the distance does not increase uniformly with time.

Q11: A train starting from rest attains a velocity of 72 km/h in 5 minutes. Calculate its acceleration and the distance covered in this time. (3 Marks)
Ans:

Q12: A big truck moving along a straight line at a speed of 54km/hr stops in 5s after the brakes are applied.

a. Find the acceleration, assuming it to be constant.

b. Plot the graph of speed versus time.

c. Using the graph. Find the distance covered by the car after the brakes are applied. (5 Marks)

Ans: 

a. Calculation of Acceleration:

First, convert the speed from km/h to m/s:

$Initial speed\left(u\right)=54km/hkm/h\; =54\times \frac{5}{18}m/s=\mathrm{15 }m/s\backslash text\{Initial\; speed\}\; (u)\; =\; 54\; \backslash ,\; \backslash text\{km/h\}\; =\; 54\; \backslash times\; \backslash frac\{5\}\{18\}\; \backslash ,\; \backslash text\{m/s\}\; =\; 15\; \backslash ,\; \backslash text\{m/s\}$m/s

The truck comes to a stop, so the final velocity $v=0$m/s.

The time is taken to stop $t=\mathrm{5 }secondst\; =\; 5\; \backslash ,\; \backslash text\{seconds\}$seconds.

The acceleration $a$a can be calculated using the first equation of motion:

$v=u+at$$\mathrm{ }$

$0=15m/s+a\times 5s0\; =\; 15\; \backslash ,\; \backslash text\{m/s\}\; +\; a\; \backslash times\; 5\; \backslash ,\; \backslash text\{s\}$ = -3m/s²

So, the acceleration of the truck is 3m/s² (negative sign indicates deceleration).

b. Plot the Graph of Speed versus Time:

The graph will be a straight line starting at 15 m/s at $t=0t\; =\; 0$ and ending at 0 m/s at $t=5t\; =\; 5$seconds. Here's a description of how to plot it:

• X-axis (Time): Mark from 0 to 5 seconds.
• Y-axis (Speed): Mark from 0 to 15 m/s.
• Plot: Start from the point (0, 15) and draw a straight line down to (5, 0).

c. Using the Graph, Find the Distance Covered by the Truck:

The distance covered can be found by calculating the area under the speed-time graph.

The graph is a right-angled triangle with:

• Base (time): 5 seconds
• Height (initial speed): 15 m/s

The area (which represents the distance) is:

$Distance=\frac{1}{2}\times Base$Base×Height$Distance=\frac{1}{2}\times 5s\times 15m/s=37.5meters\backslash text\{Distance\}\; =\; \backslash frac\{1\}\{2\}\; \backslash times\; 5\; \backslash ,\; \backslash text\{s\}\; \backslash times\; 15\; \backslash ,\; \backslash text\{m/s\}\; =\; 37.5\; \backslash ,\; \backslash text\{meters\}$ meters

So, the distance covered by the truck after the brakes are applied is $\mathrm{37.5 }meters37.5\; \backslash ,\; \backslash text\{meters\}$meters.

Q13. A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time :
The acceleration of the car is: (5 Marks)
Ans: The speed-time graph obtained from the given readings is shown here. Note that in this case, when the time is 0, then the speed is not 0. The body has an initial speed of 5 m/s, which is represented by point A.

Acceleration = slope of the speed-time graph

                   = slope of line AF = (FG/AG)

Therefore, FG = 30 - 5 = 25 m/s

Again, at point G, the value of time is 50 seconds, whereas that at point A is 0 seconds.

Thus, AG = 50 - 0 = 50 s

Now, putting these values of FG and AG in the above relation, we get :

Acceleration = = 0.5 m/s²