Advertisements
Advertisements
Question
A car starts from rest and moves along the x-axis with constant acceleration 5 ms–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?
Advertisements
Solution
For constant acceleration of car,
u = 0, a = 5 ms−2, t = 8 s
using s = `"ut" + 1/2 "at"^2`, we get
s1 = `0 + 1/2 xx 5 xx 8^2`
= 160 m
For constant velocity of car,
v = u + at
= 0 + 5 × 8
= 40 ms−1
t = (12 − 8) s = 4 s, a = 0
∴ s2 = vt
= 40 × 4
= 160 m
There fore, total distance covered in 12 s
= s1 + s2
= 160 + 160
= 320 m
APPEARS IN
RELATED QUESTIONS
Under what conditions can a body travel a certain distance and yet its resultant displacement be zero ?
If on a round trip you travel 6 km and then arrive back home :
What is your final displacement ?
The minimum distance between the start and finish points of the motion of an object is called the ______ of the object.
Can displacement-time sketch be parallel to the displacement axis? Give a reason to your answer.
From the displacement-time graph of a cyclist given below in the Figure, find The displacement from the initial position at the end of 10 s,

Write an expression for the distance S covered in time t by a body which is initially at rest and starts moving with a constant acceleration a.
Explain the following concept in your own words with everyday examples:
Distance
The table below shows the distance travelled by two vehicles A and B during each second:
| Time (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| Distance travelled by A (m) | 0 | 20 | 80 | 180 | 240 | 300 | 360 | 420 |
| Distance travelled by B (m) | 0 | 10 | 40 | 90 | 160 | 250 | 360 | 490 |
At what time do A and B meet?Which vehicle is moving with a constant acceleration?
Exercise Problems.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 m and 20 s?
Is displacement a scalar quantity?
