Motion in a Straight Line: questions with answers

Compare the vehicle's size with the distance travelled. When its size is small compared with that distance and rotation is irrelevant, a point-object approximation is appropriate. Actual size alone does not decide this.

Describing the changing orientation requires a body with size that can rotate. A point object has no orientation, even when the actual object is small.

A full eight-second cycle gains 2 m. After 5 complete cycles the next five forward steps reach the target; no subsequent backward step occurs before first arrival.

Each complete cycle advances the walker by $5-3=2$ m. The final forward leg starts at 8 m and reaches 13 m after five forward steps. Count only these forward steps from the final cycle, since the target is reached before its backward steps.

Differentiate the position law: v=2bt; the constant position term has zero derivative. Substitute t=2 s only after differentiating.

Velocity equals the slope of a position-time graph. The positive slope increases with time, so velocity increases. The plotted curve is not the particle's path.

Velocity is the slope of position against time, not the sign of position. The coordinate changes from -6 m to -2 m, so the particle moves in the positive x-direction.

Use v=u+at with both signs retained. Velocity becomes more negative, so its magnitude and therefore speed increase.

Displacement is the trapezium area under the velocity-time line. Use (initial velocity+final velocity)×time/2, including the rectangle beneath the initial velocity.

The part above the time axis contributes positive displacement; the part below contributes negative displacement. Subtract the negative triangle magnitude 2 m from the positive triangle area 8 m.

One particle has a single position at any given instant. It can occupy either listed position, but cannot occupy two different positions simultaneously. The maximum number is one.

Integrate v to obtain displacement kt³/3 over the interval. Divide by elapsed time to get kt²/3; acceleration here varies with time.

Use y=25+20t−5t² with ground at y=0. The positive root is 5 s. The lower landing level makes descent longer than ascent.

At constant deceleration b, braking distance is u²/(2b). Multiply the original braking distance by the square of the speed factor.

Use v(B relative A)=vB−vA in one common frame. Subtract A’s velocity; same-direction speeds are not added.