TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, equation of distance
x(t)=αt+βt3. . . (i)
where, α=2m∕s and β=0.01m∕s3.
Initial time, t1=2s and final time, t2=4s
As we know that,
Average velocity (vav)=‌

‌ Total displacement ‌(x)

‌ Total time ‌(t)

Since ‌‌x(t)=αt+βt3
∴ Distance travelled by car in time interval (t2−t1) will be
x‌‌=α(t2−t1)+β(t23−t13)
‌‌=2(4−2)+‌

1

100

(43−23)
‌‌=2(2)+‌

1

100

[64−8]
‌‌=4+‌

1

100

(56)
‌‌=4.56m
∴‌‌v‌av ‌‌‌=‌

x

t2−t1

=‌

4.56

4−2

‌‌=‌

4.56

2

=228m∕s