1
AP EAPCET 2024 - 21th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
The velocity of a particle is given by the equation $v(x)=3 x^2-4 x$, where $x$ is the distance covered by the particle. The expression for its acceleration is
A
$(6 x-4)$
B
$6\left(3 x^2-4 x\right)$
C
$\left(3 x^2-4 x\right)(6 x-4)$
D
$(6 x-4)^2$
2
AP EAPCET 2024 - 21th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The acceleration of a particle which moves along the positive $X$-axis varies with its position as shown in the figure. If the velocity of the particle is $0.8 \mathrm{~ms}^{-1}$ at $x=0$ , then its velocity at $x=1.4 \mathrm{~m}$ is $\left(\right.$ in $\left.\mathrm{ms}^{-1}\right)$

AP EAPCET 2024 - 21th May Evening Shift Physics - Motion in a Straight Line Question 1 English
A
1.6
B
1.2
C
1.4
D
0.8
3
AP EAPCET 2024 - 21th May Morning Shift
MCQ (Single Correct Answer)
+1
-0
An object projected upwards from the foot of a tower. The object crosses the top of the tower twice with an interval of 8 s and the object reaches foot after 16 s . The height of the tower is $\left(g=10 \mathrm{~ms}^{-2}\right)$
A
220 m
B
240 m
C
640 m
D
80 m
4
AP EAPCET 2024 - 20th May Evening Shift
MCQ (Single Correct Answer)
+1
-0
The relation between time $t$ and displacement $x$ is $t=\alpha x^2+\beta x$, where $\alpha$ and $\beta$ are constants. If $v$ is the velocity, the retardation is
A
$2 \alpha v^3 \beta^2$
B
$2 \alpha \beta v^3$
C
$-2 \beta v^3$
D
$2 \alpha v^3$
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