In this page, you can find the complete solutions of the first exercise of Kinematics: The Geometry of Motion chapter from Basic Mathematics Grade XI book published and distributed by Sukunda Pustak Bhawan.

In the above-mentioned book, Kinematics is the 22nd chapter and has three exercises only. Out of which, this is the solution of the first exercise.


Check: Basic Mathematics Grade 11 (Sukunda Publication) Guide:

Grade 11 Basic Mathematics by Sukunda Pustak Vawan Notes and Solutions | Nepal


Disclaimer:

Answers mentioned here are not solved by teachers. These are the solutions written by a student of Grade 11. Answers are all correct. However, the language or process of solving the questions might be informal and in examinations, you might have to add little more language and show more calculations than what has been done here. So, we highly encourage you to view these solutions as guide rather than just copying everything mentioned here.

Here, we have used the following notations:

u = intial velocity
a = acceleration
v = final velocity
t = time




1 i) A particle starts from rest and moves with a uniform acceleration of 5 cm/sec^2. What will be its velocity at the end of 10 seconds?

Solution:
Here,

$u = 0 cm/s$
$a = 5 cm/s^2$
$t = 10 s$
To find: final velocity (v) = ?

We know,
$v = u + at$
$= 0 + 5 * 10$
$= 50 cm/s$

1 ii) A point starts moving with initial velocity 2 cm/sec. If its acceleration is 1 cm/sec^2, find its velocity at the end of 5 seconds.

Solution:
Here,
$u = 2 cm/s$
$a = 1 cm/s^2$
$t = 5 s$

To find: final velocity (v) = ?

We know,
$v = u + at$
$= 2 + 1*5$
$= 2 +5$
$= 7 cm/s$

2 i) The velocity of a body increases from 20 m/s to 50 m/s in 10 seconds. Calculate the acceleration.

Solution:
Here,
$u = 20 m/s$
$v= 50 m/s$
$t = 10 s$
Now,
$a = \dfrac{v-u}{t}$
$= \dfrac{50-20}{10}$
$= \dfrac{30}{10}$
$= 3 m/s^2$

2 ii) Brakes are applied to a body moving with velocity 40 m/s If after describing a distance of 50m, it comes to rest. Find the retardation.

Solution:
Here,
$u = 40 m/s$
$v = 0 m/s$ [ The body finally stops. ]

$s = 50 m$
To find: retardation (-a) = ?
We have,
$v^2 = u^2 + 2(-a)s$
$or, 0^2 = 40^2 - 2 * a * 50$
$or, 100 a = 1600$
$\therefore a = 16 m/s^2$
Hence, retardation is 16 m/s-s.


3 i) A train moving with a velocity of 360 km/h has the uniform acceleration of 40 m/s^2. Obtain the distance covered by the train in 1/2 minute.

Solution:
Here,
$u = 360 km/h$
$\implies u = 360 * \dfrac{1000 m}{3600 s}$

$\implies u = 100 m/s$
And,
$a = 40 m/s^2$
$t = 1/2 minute = 30 s$
We know,
$s = ut + 1/2 at^2$
$= 100 * 30 + 1/2 * 40 * 30^2$

$= 3000 + 18000$
$= 21000 m$
$= 21 km$
3 ii) An aeroplane lands on the runway with a velocity of 108 km/h. If then its velocity slows down at the rate of 25 m/s^2, find the distance covered by the aeroplane before coming to rest.

Solution:
Here,
$u = 108 km/h$
$\implies u = 108 * \dfrac{1000m}{3600s}$

$\implies u = 30 m/s$
$v = 0 m/s$ [ Comes to rest ]
$a = -25 m/s^2$ [ Retardation ]

We have,
$v^2 = u^2 + 2as$
$or, 0^2 = 30^2 + 2(-25)s$
$or, 50 s = 30^2$
$therefore s = 18 m$

4 i) A motocar travelling with a velocity of 200 m/s has a uniform acceleartion 12 m/s^2. If its velocity at a certain interval of time is 320 m/s, find the time.

Solution:
Here,
$u = 200 m/s$
$a = 12 m/s^2$
$v = 320 m/s$
To find: t = ?
We know,
$t = \dfrac{ v- u}{a}$
$or, t = \dfrac{320 - 200}{12}$

$or, t = \dfrac{120}{12}$
$\therefore t = 10 s$

4 ii) A cyclist travelling with a velocity of 72 km/h accelerates at the rate of 4 m/s^2 until it describes a distance of 48 m. Find the time taken.

Solution:
$u = 72 km/h = 20m/s$
$a = 4 m/s^2$
$s = 48 m$
We know,
$s = ut + 1/2 at^2$
$or, 48 = 20t + 1/2 * 4 t^2$
$or, 48 = 20t + 2t^2$
$or, 24 = 10t + t^2$
$or, t^2 + 10t - 24 = 0$
$or, t^2 + 12t -2t -24 = 0$
$or, t(t + 12) - 2(t+12) = 0$
$or, (t-2) (t+12) = $
$\therefore t = 2s$ [ Time is always positive]


5) A point travelling at 20 cm/s accelerates uniformly at 5 cm/s. Find the distance travelled in the sixth second.

Solution:
Here,
$u = 20 cm/s$
$a = 5 cm/s$
We have,
$s = ut + 1/2 at^2$
To find: distance travelled in sixth second (s) = distance travelled in six seconds (s6) - distance travelled in five seconds (s5)

We know,
$s6 = u6 + 1/2 * a(6)^2$
$= 20 * 6 + 1/2 * 5 * 36$
$= 120 + 90$
$= 210 cm$
$s5 = u5 + 1/2 * a*(5)^2$
$= 20 * 5 + 1/2 * 5 * 25$
$= 100 + 62.5$
$= 162.5 cm$
Hence, $s = s6 - s5 = 210 -162.5$

$\therefore s = 47.5 cm$

6 i) A motor-cycle increases its velocity at the rate of 5 m/s^2 to 30 m/s in 4 seconds. Find its initial velocity.

Solution:
Here,
$a = 5 m/s^2$
$v = 30 m/s$

$t = 4 s$

We know,
$v = u + at$
$or, u = v - at$
$or, u = 30 - 5*4$
$or, u = 30 - 20$
$\therefore u = 10m/s$

6 ii) A bicycle slows down with a uniform retardation of 8 m/s^2 to 9 m/s after describing a distance of 35 m. Find the initial velocity of the bicycle.

Solution:
Here,
$a = 8 m/s^2$
$v = 9 m/s$
$s = 34 m$
We know, for retardation,
$v^2 = u^2 - 2as$
$or, 9^2 = u^2 -2 * 8 * 34$
$or, u^2 = 81 + 544$
$or, u^2 = 625$
$\therefore u = 25 m$ [in opposite direction of retardation]


7. A particle starts with an initial velocity 2.5 m/s along the positive x- direction and it accelerates uniformly at the rate of 0.5 m/s^2.

Solution:
$u = 2.5 m/s$
$a = 0.5 m/s^2$
i) Find the distance travelled by it in first two seconds.

When t = 2 s,
$s = ut + 1/2 at^2$
$= 2.5 * 2 + 1/2 * 0.5 * 2^2$
$= 5 + 1$

$= 6 m$

ii) How much time does it take to reach the velocity 7.5 m/s?

When v = 7.5 m/s
$v = u + at$
$or, 7.5 = 2.5 + 0.5 * t$
$or, 0.5 * t = 5$
$\therefore t = 10 s$
iii) How much distance will it cover in reaching the velocity 7.5 m/s?

When v = 7.5 m/s
$v^2 = u^2 + 2as$
$or, 7.5^2 = 2.5^2 + 2 * 0.5 *s$

$or, 56.25 = 6.25 + s$
$or, s = 56.25 - 6.25$
$\therefore s = 50 m$

8. A car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s If then runs at a constant velocity and is finally brought to rest in 64 m with a constant retardation. The total distance covered by the car is 584 m. Find the value of the aceleration, retardation and total time taken.

Solution:
Condition I,
$u = 0 m/s$
$t = 10s$
$v = 8 m/s$
$a = \dfrac{v - u}{t}$

$\implies a = \dfrac{8 - 0}{10}$

$\implies a = 0.8 m/s^2$
Also,
$v^2 = u^2 + 2as$
$or, 8^2 = 0 + 2 * 0.8 * s$
$\therefore s = 40 m$
Condition II,
When velocity is of 8 m/s, it covers distance of 64m in time 't' with a uniform retardation.

$u_1 = 8m/s$
$s_1 = 64m$
$v_1 = 0 m$
$v_1^2 = u_1^2 - 2 a_1s_1$
$or, u_1^2 = 2a_1s_1$
$or, 8^2 = 2 * a_1 * 64$
$\therefore a_1 = 0.5 m/s^2$, where $a_1$ is the retardation of the car.

$v_1 = u_1 - a_1t_1$
$or, 0 = 8 - 0.5 * t_1$
$or, 0.5 t_1 = 8$
$\therefore t_1 = 16 s$
Now,
Total distance travelled (s) = distance travelled during acceleration + distance travelled during constant velocity + distance travelled during retardation = 584 m

$or, 40 + \text{distance travlled during constant velocity} + 64 = 584$

$\therefore \text{distance travelled during constant velocity} = 480 m$

Also,
$vt_2 = 480$
$or, 8 * t_2 = 480$
$\therefore t_2 = 60$
So, the car ran at a constant velocity of 8 m/s for 60 s.

Hence, total time taken = $t + t_2 + t_1$

$= 10 + 60 + 16$

$= 86 s$


9. A point moving with uniform acceleration; in the eleventh and fifteenth second from the commencement of the motion it moves through 720 and 960 cms respectively. Find the distance covered by it in 20 sec.

Solution:
We know,
$s = ut + 1/2 at^2$
distance travelled in 11 th second = distance travelled in 11s - distance travelled in 10s

$or, 720 = ( 11 u + 1/2 * 121 a) - (10 u + 1/2 * 100a)$

$or, 720 = u + 21/2 a$ -- (i)
And,
distance travelled in 15 th second = distance travelled in 15 s - distance travelled in 14 s

$or, 960 = (15 u + 1/2 * 225 a) - (14u + 1/2 * 196 a)$

$or, 960 = u + 29/2 a$ --- (ii)

Subtracting (i) from (ii), we get,

$960 - 720 = u + 29/2 a - u - 21/2 a$

$or, 240 = 8/2 a$
$\therefore a = 60 m/s^2$
Put value of a in equation (i), we get, u = 90 m/s

And,
$t = 20 s$
$a = 60 m/s^2$
$u = 90 m/s$
Now,
$s = ut + 1/2 at^2$
$= 90 * 20 + 1/2 * 60 * 20^2$
$= 1800 + 30 * 400$
$= 1800 + 12000$
$= 13800 cm$
$= 138 m$


10. A point moving with uniform acceleration describes in the last second of its motion 9/25 of the whole distance. If it started from rest, how long was it in motion and through what distance did it move, if it described 15 cms in the first second?

Solution:

Let u be the initial velocity of the body,s  be the distance covered in t seconds, st be the distance covered in th second of its travel. let a be the acceleration. Then,

$s_t = \dfrac{9s}{25}$

$s_1 = 15 cms$

We know,

Distance travelled in 1st second,

$s_1 = u + \dfrac{1}{2} * a *(2*1 -1)$

$or, 15 = 0 + \dfrac{1}{2} * a$

$or, 15 * 2 = a$

$\therefore a = 30 cm/s^2$

And,

$s_t = u + \dfrac{1}{2} a (2t -1)$

$or, \dfrac{9s}{25} = 0 + \dfrac{1}{2} * 30 * (2t - 1)$

$or, \dfrac{9s}{25} = 15 (2t - 1)$

$or, \dfrac{3s}{25} = 5(2t - 1)$ - (i)

We know,

$s = ut + \dfrac{1}{2} at^2$

$or, s = 0 + \dfrac{1}{2} * 30 *t^2$

$or, s = 15 t^2$ - (ii)

Put value of s from (ii) in (i), we get,

$or, \dfrac{3 * (15t^2)}{25} = 5 (2t -1)$

On solving,

$or, 9t^2 -45 t - 5t +25 = 0$

$or, (9t-5)(t-5) = 0$

Either,

$9t - 5 = 0$

$\therefore t = \dfrac{5}{9}s < 1 $ [impossible]

Or,

$t - 5 = 0$

$\therefore t = 5 s$

Put value of t in equation (ii), we get,

$s = 15 * 5^2$

$= 15 * 25$

$= 375 cm$

Hence, the body was in motion for 5 seconds from the point it started to move and covered 375 cms distance in that time interval.
11. A body moves along a straight line with uniform acceleration. The body covers a distance of 18 m in the first three seconds and 22 m in the next 5 seconds.

i) What is the velocity at the end of 10 seconds?

ii) What is the distance covered in the 10th second?

Solution:
Given,

$s_{0-3} = 18 m$

$s_{3-8} = 22m$

$s_{0-8} = 18 + 22  = 40m$

Let velocity of body at 0s be u1. Clearly from the distance travelled, it can be seen that the body is retarding. Let a be the magnitude of retardation.

In time t = 3s,

$18 =  u_1*3 - \dfrac{1}{2} * a *3^2$

$or, 18 = 3u - \dfrac{1}{2} * 9a$

$or, 6 = u - \dfrac{3}{2}a$ - (i)

In time t = 8s,

$40 = u_1*8 - \dfrac{1}{2} *a *8^2$

$or, 40 = 8u - \dfrac{1}{2} * 64a$

$or, 5 = u - 4a$ (ii)

Equation equations (i) and (ii), we get,

$or, 6 + \dfrac{3}{2}a = 5 + 4a$

$\therefore a = 0.4 m/s^2$ [Remember, the body is retarding, it is just the magnitude of retardation]

Putting value of a in equation (i), we get, u = 6.6 m/s

Now,

velocity of the body at the end of 10 seconds

$v = u - 10a$

$= 6.6 - 10 * 0.4$

$= 6.6 - 4$

$= 2.6 m/s$

And,

distance covered in the 10th second is

$s = u - \dfrac{1}{2} a (2*10 -1)$

$= 6.6 - \dfrac{1}{2} * 0.4 * (19)$

$= 6.6 - 3.8$

$= 2.8 m$

Hence, the required velocity of the body at the end of 10 seconds is 2.6 m/s and it covers a distance of 2.8 m in the 10th second of its travel.



About the Textbook:

Name: Basic Mathematics Grade XI
Author(s): D.R. Bajracharya | R.M. Shresththa | M.B. Singh | Y.R. Sthapit | B.C. Bajracharya
Publisher: Sukunda Pustak Bhawan (Bhotahity, Kathmandu)
Telephone: 5320379, 5353537
Price: 695 /- (2078 BS)