Advertisements
Advertisements
प्रश्न
A body moves from rest with uniform acceleration and travels 270 m in 3 s. Find the velocity of the body at 10 s after the start.
Advertisements
उत्तर
Initial velocity u = 0 m/s
Distance travelled s = 270 m
Time taken to travel s distance = 3 s
Let 'a' be the uniform acceleration.
Using the second equation of motion,
S = ut + (1/2) at2
We get,
270 = 0 + (1/2) a (3)2
270 = 9a/2
a = 60 m/s2
Let v be the velocity of the body 10 s after the start.
Using the first equation of motion, we get
v = u + at
v = 0 + (60)(10) = 600 m/s-1
APPEARS IN
संबंधित प्रश्न
What name is given to the speed in a specified direction ?
Fill in the following blank with suitable word:
The physical quantity which gives both, the speed and direction of motion of a body is called its……………
A motorcyclist starts from rest and reaches a speed of 6 m/s after travelling with uniform acceleration for 3 s. What is his acceleration ?
If a bus travelling at 20 m/s is subjected to a steady deceleration of 5 m/s2, how long will it take to come to rest ?
A train starting from rest moves with a uniform acceleration of 0.2 m/s2 for 5 minutes. Calculate the speed acquired and the distance travelled in this time.
A car is moving on a straight road with uniform acceleration. The speed of the car varies with time as follows :
Time (s) : 0 2 4 6 8 10
Speed (m/s) : 4 8 12 16 20 24
Draw the speed-time graph by choosing a convenient scale. From this graph :
(1) Calculate the acceleration of the car.
(2) Calculate the distance travelled by the car in 10 seconds.
One of the following is not a vector quantity. This one is :
Give one example of following motion :
Variable velocity
A car accelerates at a rate of 5 m s-2. Find the increase in its velocity in 2 s.
A car is moving with a velocity 20 m s-1. The brakes are applied to retard it at a rate of 2 m s-2. What will be the velocity after 5 s of applying the brakes?
