Unit 4 MCQ Bank

If displacement is given as x = f(t), the velocity function is obtained by:

If v = f(t), acceleration is found from:

A particle is momentarily at rest when:

A particle definitely reverses direction at a time t only if:

To find total distance travelled from a position-time function over an interval with reversals, you should:

The slope of a velocity-time graph represents:

The area under a velocity-time graph between two instants represents:

The area under an acceleration-time graph over a time interval gives:

For a projectile launched with speed u at angle θ above the horizontal, the horizontal component is:

For the same launch, the vertical component is:

At the highest point of projectile motion, the vertical component of velocity is:

At the highest point of projectile motion, the acceleration is:

Neglecting air resistance, the horizontal acceleration of a projectile is:

A standard expression for maximum height H of a projectile is:

For a projectile launched upward with initial vertical component uy, the time to reach maximum height is:

If a projectile lands at the same level from which it was launched, the total time of flight is:

On level ground, horizontal range is commonly found from:

To find the time taken by a projectile to reach a wall at horizontal distance x, you should use:

After computing the time to a wall in a projectile problem, the projectile height there is found from:

If the vertical velocity vy at time t is negative, the projectile is:

At a given instant in projectile motion, the speed magnitude is:

The instantaneous direction angle α of projectile velocity relative to the horizontal is commonly found from:

On a position-time graph, a turning point occurs where:

A negative displacement value x means:

Average speed over a journey is found from:

Average velocity over a time interval is:

The equation v^2 = u^2 + 2as can be used directly only when:

If acceleration is given as a function of displacement, a = f(s), the correct relation to use is:

If a = f(t), velocity is found by:

If acceleration is given as a function of velocity, a = f(v), a useful differential form is:

The normal component of acceleration for motion along a curved path is:

Tangential acceleration is defined as:

The normal and tangential acceleration components are always:

If a body has tangential acceleration at and normal acceleration an, its total acceleration magnitude is:

If a particle moves at constant speed along a curved path, which statement is correct?

For motion along a straight path, the normal acceleration component is usually:

If θ is measured from the tangential direction to the total acceleration vector, then:

If β is measured from the normal direction to the total acceleration vector, then:

In the formula an = v^2/r, if a track diameter is given in the problem, you must first:

At any point on a smooth curved path, the velocity vector is directed:

For curved motion with changing speed, the total acceleration vector is generally:

If velocity becomes zero twice in the first 3 seconds, the total distance over that interval must be found by splitting the motion into:

When solving y = uy t - 1/2 g t^2 = 0 for a projectile returning to launch level, why do you factor out t?

Neglecting air resistance, a projectile has the same speed magnitude at two points that are:

For same-level projectile motion, the horizontal range depends directly on:

A particle with positive velocity and negative acceleration is:

Which statement is possible for a particle that moves away from the origin and then returns to it?

To decide whether a projectile is still rising at t = 2 s, the most direct quantity to inspect is:

If tangential acceleration is given as at = f(s), why is v^2 = u^2 + 2as generally not valid directly?

Normal acceleration exists because the velocity vector is changing in: