Question

Numericals based on acceleration and three equation of motion.​

Answer 1

Answer:

rd523475g2j909.;[x27-58]ld

Explanation:

Answer 2

Answer:

EQUATIONS ARE

1. V = U +AT
2. S= UT+ 1/2 AT^2
3. V^2 = U^2 +2AS

Explanation:

NUMERICALS ARE

Q. 01. A ball hits a wall horizontally at 6m/s. It rebounds horizontally at 4.4m/s. The ball is in contact with the wall for 0.04s. What is the acceleration of the ball?

Solution:

Initial velocity(u) = 6.0m/s  

Final velocity(v) = -4.4 m/s (because direction of ball has become opposite)  

time = 0.04 s  

thus, acceleration (a) = (v-u)/t  

a= [-4.4 – 6.0]/ 0.04  

a= (-10.4)/0.04  

after multiplying -10.4/0.04 by 100/100 (to make calculations simpler), we get  

a= -1040/4 = -260 m/s2  

Q.02. A cheetah is the fastest land animal and it can achieve a peak velocity of 100 km per hour up to distances less than 500 metres. If the cheetah spots his prey at a distance of 100 meters what is the minimum time it will take to get its prey?

Solution:

If the cheetah spots the prey at its top speed, the cheetah will hunt down the prey with the speed of  100 km/h. = 27.7 m/s  

Now,  time = d/s = 100/27.7 = 3.6 sec.

So, the minimum time the cheetah will take to get the prey is 3.6 s.