Q 42 An aeroplane lands at 432 km/hr and stops after covering a runaway of 4 km. Calculate (i) acceleration (ii) time in which it comes to rest.
Answers
Answer:
Given:
Initial velocity=u=432 km/h =432×5/18=120m/s
Final velocity = V= 0 m/s
Time=?
DISTANCE covered= s=4km=4×1000m
a=?
From third equation of motion,
V^2 - u^2 = 2as
0^2 - (120*120)=2*ax4000
a= - 120x120/2*4000
= -1.8 m/s2
Therefore retardation is 1.8m/s2
From FIRST equation of motion :
V=u+ at
t= V-u/a
=0-120/-1.8
=120/1.8
=66.6 secs.
Explanation:
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Given that, the initial velocity of the aeroplane is 432 km/hr, final velocity of the aeroplane is 0 km/hr and distance travelled is 4 km.
We have to find the (i) acceleration (ii) time in which aeroplane comes to rest.
i) For acceleration:
Using the Third Equation Of Motion:
v² - u² = 2as
Firstly convert km/hr into m/s and km into m.
1 km/hr = 5/18 m/s
1 km = 1000 m
Initial velocity = 432 km/hr = 120 m/s
Final velocity = 0 km/hr = 0 m/s
Distance = 4 km = 4000 m
Substitute the known values in the above formula,
→ (0)² - (120)² = 2(a)(4000)
→ -14400 = 8000a
→ -1.8 = a
(Negative sign shows retardation)
Therefore, the value of acceleration is 1.8 m/s².
ii) For time:
Using the First Equation Of Motion:
v = u + at
Substitute the known values,
→ 0 = 120 + (-1.8)t
→ -120 = -1.8t
→ 66.67 = t
Therefore, the time taken by the aeroplane is 66.67 sec.