Question

time= 20second, initial velocity =72km/hour,final velocity =36km/hour then find the length?

Answer 1

check out the answer above

Answer 2

Given that,

Time (t) = 20 s

Initial velocity (u) = 72 km/h

Final velocity (v) = 36 km/h

To find,

Distance (s) = ?

If you gotta any problem like this, always remember that, you have to find the distance. Length means distance.

Before reckoning the distance, we've to check that all the units are in their SI unit or not?

Here, initial velocity and final velocity aren't given in their SI unit.

So, we'll convert them.

As we know,

For initial velocity (u) : -

1 km/hr = 0.278 m/s

72 km/hr = 0.278 $\times$ 72

72 km/hr = 20 m/s (approx)

For final velocity (v) : -

1 km/hr = 0.278 m/s

36 km/hr = 0.278 $\times$ 36

36 km/hr = 10 m/s (approx)

So now, both the units are in their SI units. Now, we can reckon the distance.

We can compute it by using one of the following equations of motion:

(i) v = u + at

(ii) s = ut + $\frac{1}{2}$ at²

(iii) 2as = v² - u²

But for using second equation of motion, we wanna acceleration which we don't know. But we can reckon it.

For acceleration:

By using first equation of motion,

v = u + at

10 = 20 + a $\times$ 20

10 = 20 + 20a

10 - 20= 20a

- 10 = 20a

$\frac{-10}{20}$ = a

$\frac{\cancel{-10}}{\cancel{20}}$ = a

$\frac{-1}{2}$ = a

$\boxed{\bold{\pink{\mathsf{Acceleration = - 0.5\:m/s^2}}}}$

Now, we gotta the acceleration. Now, we can substitute it in the second equation of motion.

For distance:

By using second equation of motion,

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

s = $20 \times 20 + \frac{1}{2} \times - 0.5 \times (20)^2$

s = $400 + \frac{1}{2} \times - 0.5 \times 400$

s = $400 + \frac{1}{\cancel{2}} \times - 0.5 \times \cancel{400}$

s = $400 + 1 \times - 0.5 \times 200$

s = $400 + ( - 0.5) \times 200$

s = 400 + (-100)

s = 400 - 100

s = 300 m

$\boxed{\bold{\red{\mathsf{Distance = 300\:m}}}}$

Therefore, the length is 300 m.