Motion in a Straight Line — Physics STD 11 Science — Question

1. Velocity attained after time t: The velocity-time graph for positive constant acceleration of a particle is shown in the figure.

Let u be the initial velocity of the particle at t = 0 and v is the final velocity of the particle after time t. Consider two points A and B on the curve corresponding to t = 0 and t = t respectively.

Draw BD perpendicular to time axis. Also draw AC perpendicular to BD.

$\therefore$ OA = CD = u;

BC = (v - u) and OD = t

Now slope of v-t graph = acceleration (a)

$\therefore$ a = slope of v-t graph $=\tan\theta=\frac{\text{BC}}{\text{AC}}=\frac{\text{BC}}{\text{OD}}$ [$\because$ AC= OD]

$\therefore \text{a}=\frac{\text{v}-\text{u}}{\text{t}}$

$\text{v}-\text{u}=\text{at}$

 or $\text{v}=\text{u}+\text{at}$

2. Distance travelled in time t:

Let, x₀ = position of the particle at t = 0 from the origin.

x = position of the particle at t = t from the origin.

$\therefore$ (x - x₀) = S = distance travelled by the particle in the time interval (t - 0) = t

We know, distance travelled by a particle in the given time

Interval = area under velocity-time graph

$\therefore$ (x – x₀) = Area OABD (see fig, above) = Area of trapezium OABD

$=\frac{1}{2}$ [Sum of parallel sides × perpendicular distance between parallel sides]

$=\frac{1}{2}(\text{OA}+\text{BD})\times\text{AC}=\frac{1}{2}(\text{u}+\text{v})\times\text{t}$

Since $\text{v}=\text{u}+\text{at}$

$\therefore(\text{x}-\text{x}_0)=\frac{1}{2}(\text{u}+\text{u}+\text{at})\times\text{t}$

$=\frac{1}{2}(2\text{u}+\text{at})\times\text{t}=\text{ut}+\frac{1}{2}\text{at}^2$

Since $\text{x}-\text{x}_0=\text{S}$

$\therefore \text{S}=\text{ut}+\frac{1}{2}\text{at}^2$

3. Velocity attained after travelling a distance S: We know, distance travelled by a particle in time t is equal to the area under velocity-time graph. Therefore, the distance (s) travelled by a particle during time interval t is given by

S = Area under v-t graph (see fig.) or

S = area of trapenium OABD

$=\frac{1}{2}$ (sum of parallel sides) × perpendicular distance between these para

$\text{S}=\frac{1}{2}(\text{OA}+\text{BD})\times\text{ACs}\ \dots(\text{i})$

Now, acceleratiory a = slope of v- t graph

$\text{a}=\frac{\text{BC}}{\text{AC}}=\frac{\text{BD}-\text{CD}}{\text{AC}}=\frac{\text{v}-\text{u}}{\text{AC}}$

$\text{AC}=\Big(\frac{\text{v}-\text{u}}{\text{a}}\Big)\text{s}$

Also, OA = u and BD = v

Using equations (ii) and (iii) in equation (i), we get

$\text{S}=\frac{1}{2}(\text{v}+\text{u})\frac{(\text{v}-\text{u})}{\text{a}}=\frac{\text{v}^2-\text{u}^2}{2\text{a}}$

$\text{v}^2-\text{u}^2=2\text{as}$