Question

Speed of a car increases from 36 km/h to 90 km/h in 10 second find initial velocity and final velocity, acceleration and distance covered in 10 second

Answer 1

AnswEr

The distance covered is 175m,

The acceleration is 1.5m/s²

The initial and final velocity are 10m/s and 25m/s respectively

Explanation

•Given

Speed of a car increases from 36km/h to 90km/h in 10s

So ,

initial velocity ,

u = 36km/h

⇒u = 36×1000/60×60 m/s

⇒u = 36×10/36

⇒u = 10m/s

final velocity

v = 90km/h

⇒v = 90×1000/60×60

⇒v = 90×10/36

⇒v = 25m/s

and

time ,

t = 10s

•To find

initial velocity

final velocity

acceleration

distance covered

•Formula to be used

a = (v - u)/t

s = ut + (1/2)×at²

Applying the formulae on the given data

Acceleration

⇒a = (25 - 10)/10

⇒a = 15/10

⇒a = 1.5m/s²

Again distance covered

s = 10×10 + (1/2)×1.5×10²

⇒ s = 100 + 1.5×5×10

⇒s = 100 + 75

⇒s = 175m

Answer 2

Given :

Speed of a car increases from 36km/h to 90km/h in 10 seconds

To find :

• Initial Velocity
• final velocity
• acceleration and
• distance covered in 10 seconds

Formula's used:

⇒Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2as$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

⇒Conversion For speed km/hr to m/s

To convert km/hr into m/sec, multiply the number by 5 and then divide it by 18.

Solution :

Initial Velocity, u = 36km/hr

$\sf \implies u = 36 \times \frac{5}{18}$

$\sf \implies u = 5 \times 2 = 10 ms {}^{ - 1}$

Final velocity, v = 90 km/hr

$\sf \implies v = 90 \times \frac{5}{18}$

$\sf \implies v = 5 \times 5 = 25 ms {}^{ - 1}$

Now , v= 25m/s and u = 10 m/s and t = 10s

From the first equation of motion,

$\bf\:v=u+at$

$\sf \implies25 = 10 + a \times 10$

$\sf \implies25 - 10 = a \times 10$

$\sf \implies a = \dfrac{25 - 10}{10} = \dfrac{15}{10}$

$\sf\implies a = 1.5 ms {}^{ - 2}$

From the second equation of motion ,

$\bf\:s=ut+\frac{1}{2}at{}^{2}$

$\sf \implies s = 10 \times 10 + \frac{1}{2} \times 1.5 \times 10 \times 10$

$\sf \implies s = 100 + \frac{1}{2} \times \frac{15}{ \cancel{10}} \times \cancel{10} \times 10$

$\sf \implies s = 100 + 75 = 175m$

Therefore, Intital velocity =10m/s

Final velocity= 25m/s

Acceleration= 1.5m/s²

And distance covered in 10 sec = 175m

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More About this topic:

All the kinematic equations are hold good only for uniformly accelerated motion .

They are not to be used in the case of variable acceleration.