Question

A local train leaves station A; it gains speed at the rate of 1 $m{s}^{-2}$ for first 6 s and then at the rate of 1.5 $m{s}^{-2}$ until it has reached the speed of 12 m/s. The train maintains the same speed until it approaches station B; brakes are then applied, giving the train a constant deceleration and bringing it to a stop in 6 s. If the total running time of the train is 40 s. Find (a) the distance between stations A and B. (b) Draw acceleration-time, velocity-time, and the position-time relation of motion.

Answer 1

Time interval	a-t graph	v-t graph	x-t graph
0<?<6s a, = 1	Straight line parallel with time axis and above time axis.	Straight line with (+) slope passing through origin	Parabola open up and passing through origin
6s<t<$t_1$	Straight line parallel with time axis and above time axis.	Straight line with (+) slope	Parabola open up
$t_1<t<t_2$ Constant speed	Straight line parallel along time axis	Straight line parallel with time axis	Straight line with (+) slope
$t_2<t<40s$ Retardation	Straight line parallel with time axis and below time axis.	Straight line with (-) slope	Parabola open down

Analysis from acceleration-time curve:

We have change in velocity = area under acceleration-time curve.

Denoting the velocity at t = 6 s as $v_6$,

For 0 < t < 6s: $v_6$ - 0 = 6 x 1 =6m/s

For 6s < t < $t_2$: 12 - 6 = ($t_1$ - 6)(1.5)

$t_1$ = 10s
$t_2$ = 40 - 6 = 34s
For 10 s < t < 34s :Velocity is constant, acceleration is zero.

The train moves with constant velocity 12 m/s during this interval.
For 34s < t <40s: 0-12 = 6 x $a_4$
$a_4$ = -2$m{s}^{-2}$

The negative acceleration means that a-t curve lies below t-axis; and at represents decrease in velocity.

Analysis from velocity-time curve:

Area under velocity-time curve = Change in position

For 0 < t < 6s: $x_6$ - 0 = 1/2 x 6 x 6 = 18m

For 6s < t < 10s: $x_{10}-x_6$ = 1/2 x 4 (6 + 12) = 36m

For 10 s < t < 34s :$x_{34}-x_{10}$ = 24 x 12 = 288 m

For 34s < t <40s:$x_{40}-x_{34}$ = 1/2 x 6 x 12 = 36 m

On adding the change in positions, we can find the distance between stations A and B

d = $x_{40}$ - 0 = 378 m

Note: The x-t curve for constant acceleration motion is a parabola. While drawing x-t curve, remember that at any instant slope of the tangent to the curve gives the value of v at that instant.