An object initially at rest moves for t seconds with a constant acceleration a. The average speed of the object during this time interval is
An object initially at rest moves for t seconds with a constant acceleration a. The average speed of the object during this time interval is
(a) at/2 ; (b) 2a t⋅ ; (c)1/2at^2⋅ ; (d)1/2a^2t.
Ans: – Option (a) at/2.
A car starts from rest with a uniform acceleration of 4 ms–2. The distance travelled in metres at the ends of 1s, 2s, 3s and 4s are respectively,
(a) 4, 8, 16, 32 (b) 2, 8, 18, 32
(c) 2, 6, 10, 14 (d) 4, 16, 32, 64.
Ans: – Option (c).
Does the direction of velocity decide the direction of acceleration?
Ans: – Yes, the velocity direction is decided the direction of the acceleration.
Establish the relation between acceleration and distance travelled by the body.
Ans: – The relation between the acceleration and distance travelled by the body is ,
S=ut + 1/2at^2.
Where, s=total distance, u=initial velocity, t=time, and a=acceleration of motion.
Explain whether or not the following particles have acceleration:
(i) a particle moving in a straight line with constant speed, and
(ii) a particle moving on a curve with constant speed.
Ans: – There is two cases where the acceleration is present and in the other acceleration is absent. In the first case the speed is constant so the according to Newton first law acceleration will not be produced so in second case the direction is changing so acceleration will produce.
Consider the following combination of signs for velocity and acceleration of an object with respect to a one-dimensional motion along x-axis and give example from real life situation for each case:
Ans: –
| Velocity | Acceleration | Example |
| (a) Positive | Positive | Ball rolling down on a slop like slide or ramp. |
| (b) Positive | Negative | Bullet fired on water. |
| (c) Positive | Zero | Car going on road with constant speed. |
| (d) Negative | Positive | Ball falling down from upward. |
| (e) Negative | Negative | Compressing a spring. |
| (f) Negative | Zero | For going backwards with constant speed. |
| (g) Zero | Positive | When a ball thrown upward at peak velocity is zero but acceleration is 9.8m/s^2. |
| (h) Zero | Negative | When we hit a ball towards the wall at the time of striking velocity is zero and acceleration is negative. |
A car travelling initially at 7 ms–1 accelerates at the rate of 8.0 ms–2 for an interval of 2.0 s. What is its velocity at the end of the 2 s?
Ans: – As we all know, V=u + at,
Where, v=final velocity, u=initial velocity=7m/s.a=acceleration=8m/s^2, t=time=2s.
So, v=7 + 8×2= 25m/s.
A car travelling in a straight line has a velocity of 5.0 ms–1 at some instant. After 4.0 s, its velocity is 8.0 ms–1. What is its average acceleration in this time interval?
Ans: – We know, v=u + at.
Here, v=8m/s, u=5m/s, a=? T=4s.
So, average acceleration is=(8-5)/4
= 3/4=0.75m/s^2.
The velocity-time graph for an object moving along a straight line has shown in Fig. 3.32. Find the average acceleration of this object during the time interval0 to 5.0 s, 5.0 s to 15.0 s and 0 to 20.0 s.
Ans: –
The velocity of an automobile changes over a period of 8 s as shown in the table given below:
(i) Plot the velocity-time graph of motion.
(ii) Determine the distance the car travels during the first 2 s.
(iii) What distance does the car travel during the first 4 s?
(iv) What distance does the car travel during the entire 8 s?
(v) Find the slope of the line between t = 5.0 s and t = 7.0 s. What does the slope indicate?
(vi) Find the slope of the line between t = 0 s to t = 4 s. What does this slope represent?
Ans: –
The position-time data of a car is given in the table given below:
(i) Plot the position-time graph of the car.
(ii) Calculate average velocity of the car during first 10 seconds.
(iii) Calculate the average velocity between t = 10 s to t = 20 s.
(iv) Calculate the average velocity between t = 20 s and t = 25 s. What can you say about the direction of the motion of car?
Ans: –
An object is dropped from the height of 19.6 m. Draw the displacement-time graph for time when object reach the ground. Also find velocity of the object when it touches the ground.
Ans: –
An object is dropped from the height of 19.6 m. Find the distance travelled by object in last second of its journey.
Ans: – we know, s= ut + ½ at^2;
Here, a= 9.8; s=19.6; u=0;
So, 19.6 = 0 + 1/2× 9.8× t^2;
Or, t=2s.
Distance travelled in n second be, S(n) = u + a/2(2n -1).
Here, n=2;
So, S(2)= 0 + 9.8/2 × ( 4-1) = 14.7m.
Show that for a uniformly accelerated motion starting from velocity u and acquiring velocity v has average velocity equal to arithmetic mean of the initial(u) and final velocity (v).
Ans: – Initial velocity=u, final velocity=v,
Let acceleration=a, time=t.
The arithmetic means of initial and final velocity=(v+u)/2.
We know, v=u + at.
Or, t=(v-u)/a.
Here, s=ut + 1/2at^2.
Or, s=(u-v)/a ×((u+v)/2).
Average velocity=s/t=u+ v/ 2.
So, we clearly prove it that average velocity is equal to the arithmetic mean of initial and final velocity.
Find the distance, average speed, displacement, average velocity and acceleration of the object whose motion is shown in the graph.
Ans: –
A body accelerates from rest and attains a velocity of 10 ms-1 in 5 s. What isits acceleration?
Ans: – we know that, v=u + at.
Here, v=final velocity=10m/s, u= initial velocity, a=?,
T=time= 5s.
So, 10= 0 + 5a.
Or, a=10/5=2m/s^2.
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