Given figure is Time -velocity graph that indicate acceleration .
Now, according to problem
1) A graph line is parallel to time axis i.e. velocity is not changing with passes of time , so acceleration (a) = 0.
Here,
acceleration (a)= 0
time (t)= 20 sec
starting speed (u)= 40 m /s
final speed (v)= 40m /s
distance covered( s)= ?
Now using formula for linear motion ,we get
[tex]\begin{gathered} s=ut+\text{ }\frac{1}{2}at^2; \\ s=\text{ 40}\times20+\text{ 0}\begin{cases}a={0} \\ t={20}\end{cases} \\ s=800m; \end{gathered}[/tex]Answer is 800m
2) For graph line B---
Starting speed(u)= 25 m/s ( when time =0)
final speed (v) = 50 m/s (when time =20 s)
time (t)= 20 sec
acceleration (a)= ?
distance travelled (s)= ?
Now acceleration is given by
[tex]\begin{gathered} a=\frac{v-u}{t}; \\ a=\frac{50-25}{20}=\text{ }\frac{25}{20}=\text{ 1.25ms}^{-2} \end{gathered}[/tex]Again distance travelled in 20 s is given by
[tex]\begin{gathered} s=ut\text{ +}\frac{1}{2}at^2; \\ s=\text{ 25}\times20+\frac{1}{2}\times1.25\times20^2; \\ s=500+250=750\text{ m} \end{gathered}[/tex]Answer is a= 1.25m/s² and s= 750m
3) when t= 20 sec then distance travelled by A=800m and distance travelled by B= 750 m . Therefore A is ahead of B
4) distance travelled by car A in 40 sec is given by
[tex]\begin{gathered} s=ut\text{ +}\frac{1}{2}at^2 \\ s=\text{ 40}\times40+0\begin{cases}a={0} \\ u={40}\end{cases} \\ s=1600m \end{gathered}[/tex]Now distance travelled by B is given by
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