Quadratic functions

1003108303

Level: 
B
Maximum value of the quadratic function \( f \) is \( 2 \). The graph of \( f \) intersects the \( x \)-axis at the points \( [-1;0] \) and \( [3;0] \). Find the function \( f \).
\( f(x)=-\frac12x^2+x+\frac32 \)
\( f(x)=x^2-2x+3 \)
\( f(x)=x^2-2x-3 \)
\( f(x)=-\frac12x^2-x+\frac32 \)

1003108302

Level: 
B
The graph of the quadratic function \( f \) is a parabola with the vertex \( [2;5] \). The parabola intersects the \( y \)-axis at the point \( [0;3] \). Find the function \( f \).
\( f(x)=-\frac12(x-2)^2+5 \)
\( f(x)=-\frac12(x+2)^2+5 \)
\( f(x)=-2(x-2)^2+5 \)
\( f(x)=-2(x+2)^2+5 \)

1103148606

Level: 
C
If an object moving with an initial velocity \( v_0 \) is slowing down at a constant deceleration \( a \), then the distance \( s \) travelled while decelerating is described by the formula \( s=v_0t-\frac12at^2 \), where \( t \) is time of decelerating. Choose the graph, which could represent the dependency of the distance \( s \) on the time \( t \).

1103148605

Level: 
C
Suppose, an object, that is in rest, starts to accelerate with the constant acceleration \( a \). The distance \( s \) travelled by the object in time \( t \) is given by the formula \( s=\frac12at^2 \). You can see the graph of the distance \( s \) on the time \( t \) dependency in the picture. Find the acceleration \( a \) of the object.
\( 8\frac{\mathrm{m}}{\mathrm{s}^2} \)
\( 16\frac{\mathrm{m}}{\mathrm{s}^2} \)
\( 4\frac{\mathrm{m}}{\mathrm{s}^2} \)
\( 2\frac{\mathrm{m}}{\mathrm{s}^2} \)

1103148604

Level: 
B
The total mechanical energy \( E \) of an object is given by the formula \( E=mgh+\frac12mv^2 \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (about \( 10\,\mathrm{m}/\mathrm{s}^2 \)), \( h \) is the height of the object above the ground and \( v \) is the velocity of that object. Suppose that an object of the fixed mass \( m \) moves horizontally in the constant height \( h \) above the ground. Choose the graph, which could represent the dependence between the total mechanical energy (\( E \)) and velocity (\( v \)) of the object.

1103148603

Level: 
C
Consider a simple circuit in which a battery of electromotive force \( U_e \) and internal resistance \( R_i \) drives a current \( I \) through an external resistor of resistance \( R \) (see figure). The external resistor could be for example an electric light, an electric heating element, or, maybe, an electric motor. The basic purpose of the circuit is to transfer energy from the battery to the external resistor, where it actually does something useful for us (e.g. lighting a light bulb, or lifting a weight). \[ \] The power \( P \) transferred to the external resistor is described by the formula \( P=U_eI-R_i I^2 \). What maximum power can be transferred to the external resistor if we have the source with \( R_i=0.25\,\Omega \) and \( U_e=20\,\mathrm{V} \)?
\( 400\,\mathrm{W} \)
\( 80\,\mathrm{W} \)
\( 40\,\mathrm{W} \)
\( 790\,\mathrm{W} \)

1003148602

Level: 
C
Consider an object thrown at an angle of \( 30^{\circ} \) above the horizontal with the initial velocity of \( 40\frac{\mathrm{m}}{\mathrm{s}} \). How long does it take for the object to reach its maximum height? \[ \] Note: The height \( y \) of an object thrown is described by the formula \( y=v_0t\sin\alpha-\frac12gt^2 \), where \( v_0 \) is the initial velocity, \( g \) is gravitational acceleration (count with the rounded value \( 10\frac{\mathrm{m}}{\mathrm{s}^2}\)), \( t \) is the time period of the object motion in seconds, and \( \alpha \) is the angle to the horizontal at which the object is thrown.
\( 2\,\mathrm{s} \)
\( 4\,\mathrm{s} \)
\( 8\,\mathrm{s} \)
\( 1\,\mathrm{s} \)

1003148601

Level: 
C
Consider an object thrown upwards from the ground with the initial velocity of \( 30\frac{\mathrm{m}}{\mathrm{s}} \). The object moves upwards with decreasing vertical velocity until it stops. Then it starts moving vertically downwards. Find the greatest height above the ground the object does reach. \[ \] Note: The vertical distance \( y \) of a thrown object is described by the equation \( y=v_0t-\frac12gt^2 \), where \( v_0 \) is the initial velocity of the thrown object, \( g \) is gravitational acceleration (count with the rounded value \( 10\frac{\mathrm{m}}{\mathrm{s}^2}\)), and \( t \) is the time period of the object motion in seconds.
\( 45\,\mathrm{m} \)
\( 135\,\mathrm{m} \)
\( 360\,\mathrm{m} \)
\( 40\,\mathrm{m} \)