Answer:
a.52.9 km/h
b.90 km
Explanation:
We are given that






Time spend on eating lunch and buying ga=15 min.
a.Total time=30+12+45+15=102 minute=
1 hour=60 minutes
Distance=



Total distance=
Average speed=
Using the formula
Average speed=
b.Total distance between the initial and final city lies along the route=90 km
Answer 1) : 62.5 km/hour is the average velocity of the train.
2) The final velocity of the car at the end of 75 m is 14.69 m/s
Explanation:
1) Displacement of the train = 100 km + 150 km = 250 km
Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours
Average velocity=
62.5 km/hour is the average velocity of the train.
2) The acceleration of the car, a= 1.2 
Distance covered by the car,s = 75 m
Initial velocity of the car ,
= 6 m/s
Final velocity of thre car ,
=?
Using third equation of motion:


The final velocity of the car at the end of 75 m is 14.69 m/s
Answer:
A velocity time graph shows the change of velocity of an object with respect ot time. If the slope of the graph is increasing in the postive region, it means that the velocity is changing, if the slope is decreasing, it means the the velocity is decreasing, but the object is moving in the same direction (positve direction).
If this slope intersects the graph at x-axis, it means that the body has 0 velocity and has become still. After that, if the line enters in the negative region, it means that its velocity is started to increases again, but the body is movinging in the opposite direction (negative direction)
Answer:
The girl will move with constant velocity
Explanation:
If after a certain time t_0 the velocity of the girl is v_0 =gt_0 and the upward force on the girl due to rope is mg ,where g is gravitational acceleration. Then the girl will move down with the constant velocity v_0 .
The girl will move with constant velocity,as explained above.
Answer:

Explanation:
The equivalent of Newton's second law for rotational motions is:

where
is the net torque applied to the object
I is the moment of inertia
is the angular acceleration
In this problem we have:
(net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)
is the moment of inertia
Solving for
, we find the angular acceleration:
