Please quickly answer these questions..... A train starts from rest. It moves through 1 km in 100 s with unifrom acceleration. What will be its speed at the end of 100 s. A car has a velocity of 10ms k-1. It acceleration at 0.2 ms-2 for half minute. Find the distance travelled during this time and final velocity of the car. A tennis ball is hit vertically upward with a velocity of 30 ms-1. It takes 3 seconds to reach the highest point. Calculate the maximum height reached by the ball. How long it will take to return ground. A car moves with uniform velocity of 40 ms -1 for 5 s. It comes to the rest in next 10 s with uniform decelaration. Find (1) decelaration (2)total distance travelled by the car.
Answers
Answer: Ans 1: To solve this problem, you need to use a combination of the kinematic equations, given by
xf=xi+vi(tf−ti)+12a(tf−ti)2
vf=vi+a(tf−ti)
v2f−v2i=2a(xf−xi) .
where xi, xf are the initial and final positions, vi,vf are the initial and final velocities, ti,tf are the initial and final times, and a is the acceleration. We take xi=0 and xf=1000 m, and vi=0 because it starts from rest, and ti=0 and tf=100 s for the time required.
You are given an interval of space and of time. The interval of space suggests you should employ the third equation. First, we factor the left side to get
(vf−vi)(vf+vi)=2a(xf−xi) .
Now we are also given a time interval, which suggests the first or second. Taking the second equation, we can find vf−vi=a(tf−ti) , and if we plug that into the above, we get
[a(tf−ti)](vf+vi)=2a(xf−xi) .
We can then divide to get
vf+vi=2xf−xitf−ti .
Since vi=0 this will just give vf . We thus get
vf=21000 m−0100 s−0
vf=21000 m100 s
which gives the final velocity as 20 m/s.
Ans 2: ✏ Initial velocityof car = 10m/s
✏ Acceleration = 0.2
✏ Time interval = 1/2min
To Find:
1.Distance
2.Final velocity
Conversation:
✏ 1/2min = 30s
Ans 3: Initial velocity, u = 30
Final Velocity at maximum height = 0
Time = 3
Height, s =?
Using the formula, and assuming that upwards is the positive direction and that the starting point is at ground level.
v^2 = u^2 +2as
0 = 30^2 +2(-9.81)s
s = 45.9 m
The highest point reached is 45.9 metres above the starting point. The time taken for the ball to reach the ground is 6 seconds after being hit, because the time taken for the ball to reach its turning point is the same as the time is taken for it to return to its initial position from the turning point.
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