Question

Kinematics question. Please note I’ve never done kinematics before so lots and lots of explanation very necessary.

Answer 1

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

`\begin{gathered} s=ut+\text{ }(1)/(2)at^2; \\ s=\text{ 40}*20+\text{ 0}\begin{cases}a={0} \\ t={20}\end{cases} \\ s=800m; \end{gathered}`

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

`\begin{gathered} a=(v-u)/(t); \\ a=(50-25)/(20)=\text{ }(25)/(20)=\text{ 1.25ms}^(-2) \end{gathered}`

Again distance travelled in 20 s is given by

`\begin{gathered} s=ut\text{ +}(1)/(2)at^2; \\ s=\text{ 25}*20+(1)/(2)*1.25*20^2; \\ s=500+250=750\text{ m} \end{gathered}`

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

`\begin{gathered} s=ut\text{ +}(1)/(2)at^2 \\ s=\text{ 40}*40+0\begin{cases}a={0} \\ u={40}\end{cases} \\ s=1600m \end{gathered}`

Now distance travelled by B is given by