Question

an airplane lands with velocity 270km/h and comes to rest after covering 1000m. calculate the retardation. pls answer clearly​

Answer 1

Answer:

$\boxed{\mathfrak{Retardation \ (a) = -36450 \ km/h^2}}$

Given:

Initial velocity (u) = 270 km/h

Final velocity (v) = 0 km/h

Distance covered (s) = 1000 m = 1 km

To Find:

Retardation (a)

Explanation:

$\sf \star \: From \ 3^{rd} \ equation \ of \ motion: \\ \boxed{ \bold{ {v}^{2} = {u}^{2} + 2as }}$

Substituting values of v, u & s in the formula, we get:

$\sf \implies {0}^{2} = {270}^{2} + 2a(1) \\ \\ \sf \implies 0 = 72900 + 2a \\ \\ \sf \implies 2a = -72900 \\ \\ \sf \implies a = -\frac{72900}{2} \\ \\ \sf \implies a = -36450 \: km/h^2$

$\therefore$

Retardation (a) = -36450 km/h²

(Negative sign denotes it's retarded motion)

Answer 2

Explanation:

Given that, an airplane lands with velocity 270km/h and comes to rest after covering 1000m.

We have to find the retardation of the aeroplane.

From above data we have u is 270 km/hr, v is 0 m/s and s is 1000 m.

Using the third equation of motion,

v² - u² = 2as

Firstly convert km/hr into m/s. To do this divide the value by 3.6. So, u is 75 m/s.

Substitute the values,

→ (0)² - (75)² = 2a(1000)

→ - 5625 = 2000a

→ a = - 2.8

Negative sign shows retardation.

Hence, the retardation of the aeroplane is 2.8 m/s².