Q: An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:\$\frac{dv}{dt} = -2.5\sqrt{v}\$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

\$According\, \, to\, question................... \\ \int _{ 6.25 }^{ 0 }{ \left( { \dfrac { { dv } }{ { \sqrt { v } } } } \right) =\, -2.5\int _{ 0 }^{ t }{ dt } } \\ \Rightarrow \left| { 2\sqrt { v } } \right| _{ 6.25 }^{ 0 }=-2.5\, t \\ \Rightarrow 2.\, \sqrt { 6.25 } =2.5\, t \\ \, \, \therefore \, \, \, t=2\, { { s } } \\ so\, the\, correct\, optuon\, is\, A\$

Q: A body travelling with uniform acceleration crosses two points \$A\$ and \$B\$ with velocities \$20\ m/sec\$ and \$30\ m/sec\$ respectively then the speed of the body at mid point of \$A\$ and \$B\$ is

Let the acceleration of the car is \$= a\$ and the distance between \$A \$ and \$B\$ is \$=d\$\$v^2-u^2=2ad\$Given,\$v=30\,m/s\$ and \$u=20\,m/s\$\$2ad=30^2-20^2\$\$ad=\dfrac{(900-400)}{2}=250\$when the car is at the mid point of \$AB\$ then let the speed of the car is \$v_1\$\$v_1^2-20^2=2a(\dfrac d2)\$\$v_1^2=ad+400=250+400=650\$therefore \$v_1\$ is \$= 25.5\,m/s\$

Q: Change the speed of \$6\ m/s\$ into \$km/h\$

Given that, Speed \$v=6\,m/s\$ Now, speed in \$km/h\$ \$ v=6\times \dfrac{18}{5} \$ \$ v=21.6\,km/h \$ Hence, the speed is \$21.6\ km/h\$

Q: The distance between two stations is \$240\$km. A train takes \$4\$ hours to cover this distance. Calculate the speed of the train.

Distance \$=240\$kmTime taken \$=4\$ hoursSpeed\$=\$ Distance covered \$/\$time taken \$=240km/4\$ hr \$=60km/h\$Speed of train \$=60km/h\$.

Q: If we denote speed by \$S\$, distance by \$D\$ and time by \$T\$ the relationship these equation is?

\$speed = \dfrac{distance}{Time}\$

Question 1

An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by:$\frac{dv}{dt} = -2.5\sqrt{v}$ where v is the instantaneous speed. The time taken by the object, to come to rest, would be:

Accepted Answer

$According\, \, to\, question................... \ \int  _{ 6.25 }^{ 0 }{ \left( { \dfrac { { dv } }{ { \sqrt { v }  } }  } \right) =\, -2.5\int  _{ 0 }^{ t }{ dt } } \ \Rightarrow \left| { 2\sqrt { v }  } \right| _{ 6.25 }^{ 0 }=-2.5\, t \ \Rightarrow 2.\, \sqrt { 6.25 } =2.5\, t \ \, \, \therefore \, \, \, t=2\, { { s } } \ so\, the\, correct\, optuon\, is\, A$

Question 2

A body travelling with uniform acceleration crosses two points $A$ and $B$ with velocities $20\ m/sec$ and $30\ m/sec$ respectively then the speed of the body at mid point of $A$ and $B$ is

Accepted Answer

Let the acceleration of the car is $= a$  and the distance between $A $ and $B$  is $=d$$v^2-u^2=2ad$Given,$v=30\,m/s$ and $u=20\,m/s$$2ad=30^2-20^2$$ad=\dfrac{(900-400)}{2}=250$when the car is at the mid point of $AB$ then let the speed of the car is $v_1$$v_1^2-20^2=2a(\dfrac d2)$$v_1^2=ad+400=250+400=650$therefore $v_1$ is $= 25.5\,m/s$

Question 3

Four cars A, B, C and D are moving on a leveled road. Their distance versus time graphs are shown in the given figure. Choose the correct statement

Accepted Answer

The slope of distance vs time graphs represent speed. B has the minimum slope and hence the minimum speed.

Question 4

Change the speed of $6\ m/s$ into $km/h$

Accepted Answer

Given that,&#10;&#10;Speed $v=6\,m/s$ &#10;&#10;Now, speed in $km/h$&#10;&#10;  $ v=6\times \dfrac{18}{5}&#10;$ &#10;&#10; $ v=21.6\,km/h $ &#10;&#10;Hence, the speed is $21.6\ km/h$

Question 5

A bus travels 54 km in 90 minutes. The speed of the bus is?

Accepted Answer

speed = distance / time           =54 * 1000 /90 * 60           =10 m/s

Question 6

What does the path of an object look like when it is in uniform motion?

Accepted Answer

In uniform motion, the object moves along a straight path i.e. the path of object is a straight line.

Question 7

The distance between two stations is $240$km. A train takes $4$ hours to cover this distance. Calculate the speed of the train.

Accepted Answer

Distance $=240$kmTime taken $=4$ hoursSpeed$=$ Distance covered $/$time taken $=240km/4$ hr                         $=60km/h$Speed of train $=60km/h$.

Question 8

Speedometer of a car measures

Accepted Answer

Speedometer is device that measure the speed of vehicle, at any instant of time i.e measure instantaneous speed.

Question 9

If we denote speed by $S$, distance by $D$ and time by $T$ the relationship these equation is?

Accepted Answer

$speed = \dfrac{distance}{Time}$

Question 10

The correct symbol to represent the speed of an object is

Accepted Answer

Velocity is rate of change of displacement with respect to time.Velocity $V=\dfrac DT$So, unit of velocity $=\dfrac ms$Hence Option : A