Question

A car accelerates uniformity from 18km/h^h to 36km/h^-1 in 5s.Calculate the acceleration and the distance covered by the car in that time

Answer 1

Answer:

Given:

Initial velocity = 18 km/h

Final velocity = 36km/h

Time = 5second

Time is given in second so we

other values convert into m/s

Initial velocity = 18×5/18 = 5m/s

Final velocity = 36×5/18 = 10m/s

Find:

Acceleration and distance travelled

Solution:

Using first equation and find the acceleration

v = u+at

where

v = Final velocity

u = intial velocity

a = acceleration

t = time

putting value s

v = u+at

10 = 5 + a5

10 - 5 = 5a

a = 5/5

a = 1m/s squared

Now find the distance

Using second equation of motion

v^2 - u^2 = 2as

10^2 - 5^2 = 2 ×1 ×s

100 - 25 = 2s

s = 75/2

s = 37.5m

Answer :

Acceleration = 1 m/s squared

Distance = 37 m

Answer 2

Explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• Initial velocity = 18km/h

• Final velocity = 36km/h

• Time = 6sec

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• The acceleration of the car

• The distance covered by the car in that time

${\green{\underline{\underline{\bold{Solution:-}}}}}$

We are given the speed = 18km/h

To convert it into m/s we have to multiply it with 5/18

So:-

$\sf 18 \times \dfrac{5}{18}$

= 5m/s

Covert 36km/hr in m/s

$\sf 36 \times \dfrac{5}{18}$

= 2 × 5

= 10m/s

We have:-

Initial velocity = 5m/s

Final velocity = 10m/s

Time = 5sec

According to the first equation of motion

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

Substituting the values:-

$\sf { \rightarrow 10= 5 + 5a}$

$\sf { \rightarrow 10-5= 5a}$

$\sf { \rightarrow \not5= \not 5a}$

$\sf { \rightarrow a= 1m/{s}^{2}}$

According to the third equation of motion

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

Substituting the values:-

$\sf { \rightarrow {10}^{2} - {5}^{2} = 2(1)s}$

$\sf { \rightarrow {100} - {25} = 2s}$

$\sf { \rightarrow {75} = 2s}$

$\sf { \rightarrow \dfrac{75}{2} = s}$

$\sf { \rightarrow s = 37.5m}$

Therefore:-

$\boxed{\sf \purple{ { \rightarrow Acceleration \: = 1m/ {s}^{2} }}}$

$\boxed{\sf \purple{ { \rightarrow Distance \: covered \: = 37.5m }}}$