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.

Q.03. A police jeep is chasing velocity of 45 km/h. A thief in another jeep moving with a velocity of 153 km/h. police fires a bullet with muzzle velocity of 180 m/s the velocity it will strike the car of the thief is?

Solution:

Given: Velocity of police jeep = 45 m/hr = 12.5 m/s Velocity of thief’s jeep = 153 km/hr = 42.5 m/s

Velocity of bullet = 180 m/s

Now, since bullet is fired from police jeep which is going at 42.5 m/s,

Therefore, velocity of bullet is (180 + 12.5) = 192. 5 m/s.

To calculate velocity with which bullet with hit the thief we use concept of relative velocity.

Therefore, we have VBT = VB - VT

Here VB is velocity of bullet and VT is velocity of thief

VBT = 192. 5m/s - 42.5 m/s = 150m/s

Q. 05. A car is travelling at 20 m/s along a road. A child runs out into the road 50 m ahead and the car driver steps on the brake pedal. What must the car’s deceleration be if the car is to stop just before it reaches the child?

Solution:

u = 20m/s ; s = 50m and v = 0m/s

Using, V^2 = u^2 + 2as

a = (v^2 – u^2)/2s

a = (0^2 - 20^2)/(2 x 50) = - 4m/s^2

Acceleration = -4 m/s2 or, retardation = 4 m/s2

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