Question

A car acceleration uniformly from 18 km/hr to 32 km/hr in 5seconds. calculate the acceleration and the distance covered by the car​

Answer 1

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Answer 2

Correct Question

A car acceleration uniformly from 18 km/hr to 36 km/hr in 5 seconds. Calculate the acceleration and the distance covered by the car​.

Given:

Initial velocity of car,u= 18 km/h

Final velocity of car,v= 36 km/h

Time taken by car,t= 5 s

To Find:

The acceleration and the distance covered by the car​

Diagram:

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Solution:

We know that,

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to third equation of motion for constant acceleration,

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

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

It is given that,

$\longrightarrow\rm{u=18\ km/h=18\times\dfrac{5}{18}\ m/s=5\ m/s}$

$\longrightarrow\rm{v=36\ km/h=36\times\dfrac{5}{18}\ m/s=10\ m/s}$

$\rule{190}{1}$

Let the acceleration of car be 'a' and distance covered by car be s  

So, on applying first equation of motion on car, we get

$\longrightarrow\rm{10=5+a(5)}$

$\longrightarrow\rm{10-5=5a}$

$\longrightarrow\rm{5=5a}$

$\longrightarrow\rm{5a=5}$

$\longrightarrow\rm{a=\dfrac{5}{5}}$

$\longrightarrow\rm\green{a=1\ m/s^{2}}$

Also, on applying third equation of motion on car, we get

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{(10)^{2}-(5)^{2}=2(1)s}$

$\longrightarrow\rm{100-25=2s}$

$\longrightarrow\rm{75=2s}$

$\longrightarrow\rm{2s=75}$

$\longrightarrow\rm{s=\dfrac{75}{2}}$

$\longrightarrow\rm\green{s=37.5\ m}$

Hence, acceleration of car is 1 m/s² and distance covered by car is 37.5 m.