Motion in a Straight Line — Practice

Question 1

A particle moves with v = 10 + 2t m/s. Find the displacement from t = 0 to t = 5 s.

Accepted Answer

Displacement is the integral of velocity: x = integral of (10 + 2t) dt = 10t + t^2, evaluated from 0 to 5 = 50 + 25 = 75 m. Note this is the calculus route of Example 2.2, which works whether or not acceleration is constant.

Question 2

A car moving at 72 km/h stops in 50 m. Find the deceleration.

Accepted Answer

u = 72 km/h = 20 m/s. Using v^2 = u^2 + 2as with v = 0: 0 = 400 + 2a(50), so a = -4 m/s^2. The negative sign means the acceleration opposes the motion, which is what makes the car slow down.

Question 3

Two 100 m trains move at 54 km/h and 72 km/h in opposite directions on parallel tracks. Find the crossing time.

Accepted Answer

Opposite directions, so the speeds add: relative speed = 54 + 72 = 126 km/h = 35 m/s. The gap to clear is both lengths, 100 + 100 = 200 m. Time = 200/35 = 5.71 s.

Question 4

A stone is dropped from an 80 m tower. Find the time to reach the ground and the impact speed (g = 10 m/s^2).

Accepted Answer

Dropped means u = 0, so s = (1/2)g t^2 gives 80 = 5 t^2 and t = 4 s. Then v = g t = 10 x 4 = 40 m/s.

Question 5

Can an object have zero velocity but non-zero acceleration? Give an example.

Accepted Answer

Yes. At the highest point of a ball thrown vertically upward the velocity is momentarily zero, but the acceleration is still g downward. Gravity never switches off during the flight — if it did, the ball would stay suspended at the top instead of falling back.

Question 6

A car's stopping distance at 15 m/s is 20 m. Assuming the same braking deceleration, estimate its stopping distance at 30 m/s, and explain the reasoning.

Accepted Answer

Stopping distance d_s = v0^2 / 2a is proportional to the SQUARE of the initial speed, so doubling the speed from 15 to 30 m/s multiplies the distance by four: about 80 m. This square dependence, not the linear one people intuitively expect, is why speed limits matter so much for road safety.

Question 7

A body falls from rest. Show that the distances it covers in successive equal intervals of time are in the ratio 1 : 3 : 5 : 7.

Accepted Answer

With u = 0, distance after n intervals of length tau is y_n = (1/2)g(n*tau)^2, proportional to n^2. The distance covered during the n-th interval alone is y_n - y_(n-1), proportional to n^2 - (n-1)^2 = 2n - 1. Running n = 1, 2, 3, 4 gives 1, 3, 5, 7. This is Galileo's law of odd numbers, and it follows purely from distance growing as t^2.

Question 8

On a velocity-time graph, what do the slope and the area under the curve each represent, and why is the area a displacement rather than a distance?

Accepted Answer

The slope is acceleration (dv/dt) and the area is displacement. It is a displacement because the area is a signed quantity: wherever velocity is negative the contribution counts negative, so opposite motions cancel exactly as displacement requires. Distance would need the magnitude of each interval's area summed separately, with no cancellation allowed.

Practice — Motion in a Straight Line

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