a A body starts sliding over a horizontal surface with an initial velocity of 0.5 m/s. Due to friction, its velocity
decreases at the rate of 0.05 m/s. How much time will it take to stop?
b. The brakes applied to a car produce a negative acceleration of 6m/s. If the car takes 2 seconds to stop after
applying the brakes, calculate the distance it travels during this time.
c. A body moving with a velocity of 6m/s uniformly accelerates at 1.5m/s. What will be its velocity after 5
seconds and how far will it move during this time?
d. An object of mass 20 kg is accelerated uniformly from a velocity of 36km/h to 54km/h in 25 secs. Calculate
the initial and final momenta of the object. Also find the magnitude of the force exerted on the object.
e. A girl of mass 50 kg jumps out of a boat of mass 300 kg with horizontal velocity of 3m/s. With what velocity
does the boat begin to move backwards?
pls give all the answer with step - by - step explanations.......
Answers
Answer:
a) u=0.5 m/s
a=-0.05 m/s²
v=0
By first equation of motion, we have:
v=u+at
0=0.5-0.05t
0-0.5=-0.05t
-0.5=-0.05t
0.5/0.05=t
50/5=t
10 s=t
b)Acceleration a = -6 m/s^2
Time t = 2 s
Final velocity v = 0 m/s
Let initial velocity be u
Let distance be s
v = u + at
So, 0 = u + (-6)(2)
So, u = 12 m/s
Now, s = ut + (1/2) at^2
So, s = 12(2) + (1/2)(-6)(2^2)
So, s = 24 - 12
So, s = 12 m
Thus, distance travelled is 12 m
c) GIVEN-
u=6 m/s
t=5s
a=1.5 m/s^2
v=?
v=u+at
v=6+7.5=13.5 m/s
Now,s=ut+1/2at^2
s=30+3.75
s=33.75 m/s.
d) Initial momentum = 20 kg x 36 km/h x (1000/3600) m/s = 200 kg m/s
Final momentum = 20 kg x 54 km/h x (1000/3600) m/s = 300 kg m/s
Change in momentum = 300 - 200 = 100 kg m/s
Force exerted = change in momentum/time
Force = 100 kg m/s / 25 s
Force = 4 kg m/s^2
Force = 4 N