- A$\mathrm{km} / \mathrm{s}^2$
- B$\mathrm{cms}^{-2}$
- ✓$\mathrm{km} / \mathrm{s}$
- D$\mathrm{m} / \mathrm{s}^2$
26 questions · timed · auto-graded
Speed is the distance travelled by a moving body per unit time.
Acceleration of the moving object is $4 \mathrm{~m} / \mathrm{s}^2$
Initial velocity is $20 \mathrm{~m} / \mathrm{s}$.
Time taken is $2 \sec .$
By applying $1^{\text {st }}$ equation of motion we get,
$v=u+a t$
$=20+2(4)$
$=28 \mathrm{~m} / \mathrm{s}$
It is given that the velocity time curve is parallel to $x$ axis hence velocity is constant with time and acceleration is $0$.
We are given with the relationship that,
$x=k t^2$
Differentiating it twice we get,
$\mathrm{v}=2 \mathrm{kt}$
$\mathrm{a}=2 \mathrm{k}$
Where,
$\mathrm{v} \text { - velocity }$
$a -$ acceleration
Final velocity $=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s}$
Initial velocity is $36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s}$
Time taken is $10 \sec $. So,
Acceleration $=\frac{20-10}{10}$
$=1\text{m/s}^2$
The area under the speed time graph gives us the distance.
We have,
Initial velocity $= 20\ m/s$
Final velocity $= 0\ m/s$
Time taken $= 5 \sec.$
So,
Deceleration $=$ -Acceleration
$= \frac{-(0 - 20)}{5}$
$= 4\text{m/s}^2$