Advertisements
Advertisements
प्रश्न
A car acquire a velocity of 72 km per hour in 10 second starting from rest. Find
(1) the acceleration,
(2) the average velocity, and
(3) the distance travelled in this time.
Advertisements
उत्तर
We have the following information,
Initial velocity, (u) = 0 m/s
Final velocity
`(v)` = 72 km/hr
= `"72(1000)"/3600` m/s
= 20 m/s
Time taken , (t) = (10) s
(i) So acceleration,
`a = (v-u)/t`
Put the values in the above equation to get the value of acceleration
`a = (20-0)/10`
= 2 `"m/s"^2`
(ii) we have to find the average velocity. we will use the following relation,
Average velocity = `"Initial velocity + Final velocity"/2`
So , Average velocity = `(v+u)/2`
Therefore putting the value in the above to get the avearge velocity ,
Average velocity = `(20+0)/2` m/s
= 10 m/s
(iii) We have to calculate the distance travelled. We will use the relation,
Distance travelled = (Average velocity) (Time)
So distance travelled is,
= (10)(10) m
= 100 m
APPEARS IN
संबंधित प्रश्न
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h−1 in 10 minutes. Find its acceleration.
Describe the motion of a body which is accelerating at a constant rate of 10 m s–2. If the body starts from rest, how much distance will it cover in 2 s ?
Derive the formula s= `ut+1/2at^2` , where the symbols have usual meanings.
A boy is sitting on a merry-go-round which is moving with a constant speed of 10 m s−1. This means that the boy is :
A bus increases its speed from 36 km/h to 72 km/h in 10 seconds. Its acceleration is :
State how the velocity-time graph can be used to find
The acceleration of a body
How can you find the following?
Velocity from acceleration – time graph.
Area under a v – t graph represents a physical quantity which has the unit
A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement–time graph is shown in Fig.8.4. Plot a velocity-time graph for the same.

Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of `"u"_1^2 : "u"_2^2` (Assume upward acceleration is –g and downward acceleration to be +g)
