114 - Kinematics Questions Answers
A rocket is fired vertically from the ground. It moves upward with a constant acceleration 10 m/s2 for 30s after which the fuel is consumed. After what time from the instant of firing the rocket will attain the maximum height?
total height covered in upward journey = at2/2 = 10*30*30/2 = 4500 = 4.5 km
after consuming all the fuel, it will fall with gravity and g = 10 m/s2 so it will move the same distance again upto stop
so max H = 9 km.
a bullet is fired vertically upwards with an initial velocity of 50 m/s. It covers a distance h1 during the first second and a distance h2 during the last 3 seconds of its upward motion. If g = 10 m/s2, h1 and h2 will be related as
remember that gravity reduces velocity by 10 m/s on upward journey.
Also understand the diagram given
a car is going eastwards with a velocity of 8m/s. To the passengers in the car , a train appears to be moving northwards with a velocity 15m/s. What is the actual velocity of the train?
vobserver = 8i
vtrain,observer = 15j
so by formula
vtrain,observer = vtrain - vobserver
The displacement x of a particle varies with time t as x2= 1+t2 . What is its acceleration?
on diffrentiation we will get
2xdx/dt = 2t
so xv = t (1)
on differentiating again
xdv/dt + vdx/dt = 1
or xa + v2 = 1 (2)
now put v from (1) to (2) we will get answer
a particle moving with constant acceleration on a straight line ultimately comes to rest. What is the angle between its initial velocity and acceleration
180 degree because it is retarded particle
a particle moves along a straight line. Its position (x) at any instant is given by x= 32t - 8t3/3, where x is in metre and t in second. Find the acceleration of the particle at the instant particle comes to rest
differentiate the given function and find v, again differentiate find a
compare v to 0 for rest, get t and put in a
a particle initially at rest moves along x-axis. Its acceleration is given as a = (2t+2) m/s2. If it starts from origin , find distance covered in 2 second.
dv/dt = 2t+2
so dv = 2tdt+2dt
on integrating the given eq.
v = t2 + 2t + C
at start v = 0 at t = 0
so on puting these initial values we get C = 0
so v = t2 + 2t
so dx/dt = t2 + 2t
so dx = t2dt + 2tdt
on integrating x = t3/3 + t2+ C
on putting initial values x=0 at t=0
we get C = 0
so x = t3/3 + t2
now put t = 2
Angle between a and b is θ. What is the value of a.(bxa) ?
If two vectors are some in scalar triple product then it will be 0 always
the velocity v of a particle moving in the positive direction of x - axis as v = 5√x, assuming that at t=0, particle was at x=0. What is the acceleration of the particle?
v = 5√x
differetiate this we get
dv/dt = 5*1/(2√x)*dx/dt
so a = 5*1/(2√x)*v
so a = 5*1/(2√x)*5√x
so a = 12.5
a man wants to cross a river of width 500m in minimum distance. The rowing a speed of man relative to water is 3 km/h and river flows with speed 2km/h. The time taken by man to cross the river is
for minimum drift he should swim in a direction such that 3sinA = 2
so sinA = 2/3
now calculate cosA
then time = 500/3cosA