Question

A car having initial velocity 8 m/s moves with a constant acceleration of 8 m/s2 on a straight road. The ratio of distance travelled by car in first, third and fifth second of its motion is

Answer 1

Initial velocity=u=8m/s

Acceleration=a=8m/s^2

Using second equation of kinematics

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

s=distance

t=1s:-

$\\ \sf\longmapsto s=8(1)+\dfrac{1}{2}\times 8\times (1)^2$

$\\ \sf\longmapsto s=8+4$

$\\ \bf\longmapsto s_1=12m$

t=3s:-

$\\ \sf\longmapsto s=8(3)+\dfrac{1}{2}\times 8\times (3)^2$

$\\ \sf\longmapsto s=24+4(9)$

$\\ \sf\longmapsto s=24+36$

$\\ \bf\longmapsto s_2=60m$

t=5s:-

$\\ \sf\longmapsto s=8(5)+\dfrac{1}{2}\times 8\times (5)^2$

$\\ \sf\longmapsto s=40+8(25)$

$\\ \sf\longmapsto s=40+200$

$\\ \bf\longmapsto s_3=240m$

Ratio:-

$\\ \sf\longmapsto \dfrac{s_1}{\dfrac{s_2}{s_3}}=\dfrac{12}{\dfrac{60}{240}}$

$\\ \sf\longmapsto \dfrac{s_1}{\dfrac{s_2}{s_3}}=\dfrac{1}{5}{20}$

$\\ \bf\longmapsto s_1:s_2:s_3=1:5:20$

Answer 2

Initial Velocity is the velocity at time interval t = 0 and it is represented by u. It is the velocity at which the motion starts. They are four initial velocity formulas: (1) If time, acceleration and final velocity are provided, the initial velocity is articulated as. $u = v - at.$

Distance is a numerical measurement of how far apart objects or points are. In Physics or everyday usage, distance may refer to a physical length or an estimation based on other criteria. The distance from a point A to a point B is sometimes denoted as |AB|.

In mechanics, acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

Initial velocity$=u=8m/s$

Acceleration$=a=8m/s^2$

Using second equation of kinematics

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

s=distance

t=1s:-

$\rightarrow s=8(1)+\frac{1}{2} \times 8\times (1)^2$

$\rightarrow s=8+4$

$\rightarrow s_1=12m$

t=3s:-

$\rightarrow s=8(3)+\frac{1}{2}\times 8 \times (3)^2$

$\rightarrow s=24 +4(9)$

$\rightarrow s=24+36$

$\rightarrow s_2 =60m$

t=5s:-

$\rightarrow s=8(5)+\frac{1}{2}\times 8 \times (5)^2$

$\rightarrow s=40 +8(25)$

$\rightarrow s=40+200$

$\rightarrow s_3 =240m$

Ratio:-

$\rightarrow \frac{s_1}{s_2}=\frac{12}{60}$

$\rightarrow \frac{s_1}{s_2} =\frac{1}{5}20$

$\rightarrow s_1 : s_2 : s_3 =1:5:20$

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