Question

A car starting from rest atains a
velocity of 60 km/h in 10 min. Calculate the acceleration and distance traveled by The care, assuming that acceleration is uniform​

Answer 1

$\large\frak{given} \begin {cases} \: \sf initial \: velocity (u)= \frak{0} km/hr \\ \sf \: final \: velocity(v) = \frak{60}km/hr \\ \sf \: time(t) = \frak{10 \: min = \red{\dfrac{1}{6} hr}}\end{cases}$

• Intial Velocity is 0km/hr because the vehicle is starting from the rest.
• We are provided with above information from the question and we are asked to find the acceleration of this car and distance traveled by this car.
• So, First we have to calculate the value of acceleration by using 1st equation of motion.

$: \implies \underline{ \boxed { \pink{{\frak{v = u + at}}}}}$

$\: \: \: \: \: \: \: \dag \: \underline \frak{substituting \: the \: given \: values \: in \: the \: equation : - }$

$\: \: \: \: \: \: \: : \implies \bf \: 60 = 0 + a \bigg(\dfrac{1}{6} \bigg)$

$\: \: \: \: \: \: \: : \implies \bf 60 = \dfrac{a}{6}$

$\: \: \: \: \: \: \: : \implies \bf a = 6 \times 60$

$\: \: \: \: \: \: \: : \implies {\boxed{ \pink{\frak{a = 360 m/ {s}^{2} }}}} \bigstar$

• Now, we are going to find the distance traveled by this car with the help of 3rd equation of motion.
• you can use 2nd equation as well... there will be no change in the answer but for the sake of my simplicity I'm using 3rd equation :)

$: \implies \underline{ \boxed { \pink{{\frak{v²=u²+2as}}}}}$

• the letter "s" introduced in this equation represents "distance traveled by the body"

$\: \: \: \: \: \: \: \dag \: \underline \frak{substituting \: the \: given \: values \: in \: the \: equation : - }$

$\: \: \: \: \: \: \: : \implies \bf \: {(60)}^{2} = {(0)}^{2} + 2(360)s$

$\: \: \: \: \: \: \: : \implies \bf \: 3600 = 0 + 720s$

$\: \: \: \: \: \: \: : \implies \bf \: s = \cancel\dfrac{3600}{720}$

$\: \: \: \: \: \: \: : \implies {\boxed{ \pink{\frak{s = 5 \: m }}}} \bigstar$

$\therefore {\underline{ \sf {our \: respective \: answers \: are \: 360m {s}^{ - 2} \: and \: 5 \: m}}}$

i hope it's beneficial :D

Answer 2

Given :-

A car starting from rest atains a velocity of 60 km/h in 10 min. And the acceleration is uniform .

To Find :-

The acceleration atains by the car and distance covered by the car .

Used Concepts :-

• S.I Unit of acceleration is m/s².
• S.I unit of distance is metre ( m ).
• There are three formulae related to velocity , acceleration and distance i.e :-

1. v² - u² = 2as
2. s = ut + 1/2at²
3. v = u + at

Here , v = final velocity , u = initial velocity , s = distance covered , a = acceleration and t = time taken.

• We can use any formulae as the given situation.
• If a car is starting from rest then , the initial velocity is "0".
• S.I unit of velocity is m/s.
• a = v - u /t

Solution :-

Here , u = 0 m/s

v = 60 km/h

Here , the final velocity is in km/h but , we have to convert it in m/s.

=> 60 × 1000 m/h

=> 60000 m/h

=> 100/6 = 50/3 m/s.

t = 10 min = 10 × 60 = 600 s.

Now. ,As we know that ,

a = v - u / t

$a = \frac{ \frac{50}{3} - 0}{600}$

$a = \frac{ \frac{50}{3} }{600}$

$a = \frac{50}{3} \times \frac{1}{600}$

$a = \frac{5}{180}$

Therefore , acceleration of the car is 5/180 m/s.

Now , Distance ( s ) = ?

s = ut + 1/2 at²

s = 0 × 600 + 1 × 5 × 600 × 600/2 × 180

s = 5 × 1000

s = 5000 m

s = 5000/1000 km

s = 5km .

Henceforth , The acceleration and distance covered by the car is 5/180 m/s² and 5000m respectively.