Question

please answer it fast it's urgent
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Answer 1

Given:-

→ Initial velocity of the car (u) = 30km/h

→ Final velocity of the car (v) = 60km/h

→ Time (t) = 15 sec

To find:-

→ Distance travelled in given time

Solution:-

Firstly let's convert the initial and final velocity of the car from km/h to m/s.

=> 1km/h = 5/18m/s

=> 30km/h = 30×5/18

=> 75/9 m/s (u)

=> 60km/h = 60×5/18

=> 150/9 m/s (v)

At next step, we have to find the acceleration of the car using the

1st equation of motion.

=> v = u + at

=> 150/9 = 75/9 + a(15)

=> 150/9 - 75/9 = 15a

=> 75/9 = 15a

=> a = 75/[9×15]

=> a = 5/9 m/s²

Now, let's calculate the required distance by using the 2nd equation of motion.

=> s = ut + 1/2at²

=> s = 75/9 × 15 + 1/2×5/9×(15)²

=> s = 125 + 5/18 × 225

=> s = 125 + 62.5

=> s = 187.5m

Thus, distance travelled by the car in

15 seconds is 187.5m .

Answer 2

Explanation:

$\sf Given\begin {cases}initial \:velocity_(u)=30km/h \\ final \:velocity_(v)=60km/h \\ time\:taken_(t)=15s\end {cases}$

To find:-

Distance travelled=s

Solution:-

• First we need to convert km/h into m/s

$\longmapsto$$\sf {30km/h=30×{\dfrac {1000}{3600}}} \\ \sf{={\cancel {30}}×{\dfrac {5}{{\cancel{18}}}}} = {\dfrac {25}{3}}m/s$

$\longmapsto$$\sf {60km/h={\cancel {60}}×{\dfrac {5}{{\cancel {18}}}}} \\ \sf{={\dfrac {50}{3}}m/s}$

• Using first equation of kinematics

${\boxed{\sf {v=u+at}}}$

• Substitute the values

$\longmapsto$$\sf {{\dfrac {50}{3}}={\dfrac {25}{3}}+15a}$

$\longmapsto$$\sf{{\dfrac {50}{3}}-{\dfrac {25}{3}}=15a }$

$\longmapsto$$\sf {{\dfrac{50-25}{3}}=15a}$

$\longmapsto$$\sf {15a ={\dfrac {25}{3}}}$

$\longmapsto$$\sf {45a=25}$

$\longmapsto$$\sf {a={\dfrac {25}{45}}}$

$\longmapsto$${\underline{\boxed{\bf {a={\dfrac {5}{9}}m/s{}^{2}}}}}$

• Now use 3rd equation of kinematics

${\boxed {\sf {v {}^{2}-{u}^{2}=2as}}}$

• Substitute the values

$\longmapsto$$\sf {({\dfrac {50}{3}}){}^{2}-({\dfrac {25}{3}}){}^{2}=2×{\dfrac {5}{9}}×s}$

Use algebraic identitiy

${\boxed{\sf{(a+b)(a-b)={a}^{2}-{b}^{2}}}}$

$\longmapsto$$\sf {({\dfrac {50}{3}}+{\dfrac {25}{3}})({\dfrac {50}{3}}-{\dfrac {25}{3}})=2×{\dfrac {5}{9}}×s}$

$\longmapsto$$\sf {({\dfrac {50+25}{3}})({\dfrac {50-25}{3}})={\dfrac {10}{9}}s}$

$\longmapsto$$\sf {{\dfrac {75}{3}}×{\dfrac {25}{3}}={\dfrac {10}{9}}s}$

$\longmapsto$$\sf {{\dfrac{1875}{6}}={\dfrac {10}{9}}×s}$

$\longmapsto$$\sf{s={\dfrac{1875}{6}}×{\dfrac {9}{10}}}$

$\longmapsto$$\sf {s={\dfrac {1875×9}{6×10}}}$

$\longmapsto$$\sf {s={\dfrac{16875}{60}}}$

$\longrightarrow$$\sf {s=281.25m}$

$\therefore$$\sf {Distance\:travelled\:by\:the\:car\:during\:the\:period} \\ \sf {of\;15seconds\;is{\quad}281.25m.}$