Describing Motion Around Us — CBSE Class 9 Science Important Questions

Answer: (b) $\text{m s}^{-2}$m s^-2.

Acceleration is change of velocity ($\text{m s}^{-1}$m s^-1) per unit time ($\text{s}$s), so its unit is $\dfrac{\text{m s}^{-1}}{\text{s}}=\text{m s}^{-2}$m s^-1s=m s^-2.

Answer: (c) Zero.

After one complete round the body returns to its starting point, so the shortest distance between initial and final positions (displacement) is zero, even though the distance travelled equals the circumference.

Answer: (a) $4\ \text{m s}^{-2}$4 m s^-2.

Using $a=\dfrac{v-u}{t}=\dfrac{20-0}{5}=4\ \text{m s}^{-2}$a=v-u/t=20-0/5=4 m s^-2.

Answer: (c) Distance (displacement) travelled.

On a velocity-time graph, velocity $\times$× time gives displacement; geometrically this is the area between the line and the time axis. The slope, not the area, gives acceleration.

Answer: (b) Both A and R are true but R is not the correct explanation of A. Displacement can be zero when the body returns to its start (as in a round trip), but the reason for this is the return to the initial position, not merely that displacement is a vector.

Speed is the distance travelled per unit time; it is a scalar (only magnitude): $\text{speed}=\dfrac{\text{distance}}{\text{time}}$speed=distance/time.

Velocity is the displacement per unit time; it is a vector (magnitude and direction): $\text{velocity}=\dfrac{\text{displacement}}{\text{time}}$velocity=displacement/time. A body can have constant speed yet changing velocity if its direction changes.

Convert: $u=54\ \text{km h}^{-1}=54\times\dfrac{5}{18}=15\ \text{m s}^{-1}$u=54 km h^-1=54×5/18=15 m s^-1, $v=0$v=0, $t=10\ \text{s}$t=10 s.

$a=\dfrac{v-u}{t}=\dfrac{0-15}{10}=-1.5\ \text{m s}^{-2}$a=v-u/t=0-15/10=-1.5 m s^-2.

The negative sign shows retardation, so acceleration $=-1.5\ \text{m s}^{-2}$=-1.5 m s^-2.

Let a body have initial velocity $u$u and uniform acceleration $a$a. After time $t$t its velocity is $v$v.

By definition, acceleration is the rate of change of velocity:
$a=\dfrac{v-u}{t}$a=v-u/t

Multiplying both sides by $t$t:
$at=v-u$at=v-u

Rearranging gives the first equation of motion:
$v=u+at$v=u+at

Given $u=0$u=0, $v=72\ \text{km h}^{-1}=72\times\dfrac{5}{18}=20\ \text{m s}^{-1}$v=72 km h^-1=72×5/18=20 m s^-1, $t=5\ \text{min}=300\ \text{s}$t=5 min=300 s.

(i) $a=\dfrac{v-u}{t}=\dfrac{20-0}{300}=\dfrac{1}{15}\approx0.067\ \text{m s}^{-2}$a=v-u/t=20-0/300=1/150.067 m s^-2.

(ii) $s=ut+\tfrac12at^2=0+\tfrac12\times\dfrac{1}{15}\times(300)^2=\dfrac{90000}{30}=3000\ \text{m}=3\ \text{km}$s=ut+12at^2=0+12×1/15×(300)^2=90000/30=3000 m=3 km.

Taking upward as positive, $u=19.6\ \text{m s}^{-1}$u=19.6 m s^-1, $a=-g=-9.8\ \text{m s}^{-2}$a=-g=-9.8 m s^-2. At the highest point $v=0$v=0.

Maximum height using $v^2=u^2+2as$v^2=u^2+2as:
$0=(19.6)^2+2(-9.8)s\implies s=\dfrac{384.16}{19.6}=19.6\ \text{m}.$0=(19.6)^2+2(-9.8)s s=384.16/19.6=19.6 m.

Time using $v=u+at$v=u+at:
$0=19.6-9.8t\implies t=\dfrac{19.6}{9.8}=2\ \text{s}.$0=19.6-9.8t t=19.6/9.8=2 s.

(i) Acceleration (0 to 4 s): velocity rises from $0$0 to $20\ \text{m s}^{-1}$20 m s^-1 in $4\ \text{s}$4 s.
$a_1=\dfrac{20-0}{4}=5\ \text{m s}^{-2}.$a_1=20-0/4=5 m s^-2.

(ii) Retardation (8 to 10 s): velocity falls from $20\ \text{m s}^{-1}$20 m s^-1 to $0$0 in $2\ \text{s}$2 s.
$a_3=\dfrac{0-20}{2}=-10\ \text{m s}^{-2}\ (\text{retardation }10\ \text{m s}^{-2}).$a_3=0-20/2=-10 m s^-2 (retardation 10 m s^-2).

(iii) Total distance = area under the graph (a trapezium):
Parallel sides are the constant-velocity time $=4\ \text{s}$=4 s and the total time $=10\ \text{s}$=10 s, height $=20\ \text{m s}^{-1}$=20 m s^-1.
$s=\tfrac12(4+10)\times20=\tfrac12\times14\times20=140\ \text{m}.$s=12(4+10)×20=12×14×20=140 m.

Derivation: On a velocity-time graph for uniform acceleration, the line rises from $u$u (at $t=0$t=0) to $v$v (at time $t$t). The displacement equals the area under the line, which is a trapezium made of a rectangle (height $u$u, width $t$t) and a triangle (base $t$t, height $v-u=at$v-u=at).

$s=\underbrace{u\times t}_{\text{rectangle}}+\underbrace{\tfrac12\times t\times at}_{\text{triangle}}$s=u× t_rectangle+12× t× at_triangle
$\boxed{s=ut+\tfrac12at^2}$s=ut+12at^2

Numerical: $u=2\ \text{m s}^{-1}$u=2 m s^-1, $a=3\ \text{m s}^{-2}$a=3 m s^-2, $t=4\ \text{s}$t=4 s.
$s=2\times4+\tfrac12\times3\times4^2=8+24=32\ \text{m}.$s=2×4+12×3×4^2=8+24=32 m.

Total time $=200+100=300\ \text{s}$=200+100=300 s.

(i) Total distance $=600+200=800\ \text{m}$=600+200=800 m.

(ii) Taking east as positive, displacement $=600-200=400\ \text{m}$=600-200=400 m due east.

(iii) Average speed $=\dfrac{\text{total distance}}{\text{total time}}=\dfrac{800}{300}\approx2.67\ \text{m s}^{-1}$=total distance/total time=800/3002.67 m s^-1.

(iv) Average velocity $=\dfrac{\text{displacement}}{\text{total time}}=\dfrac{400}{300}\approx1.33\ \text{m s}^{-1}$=displacement/total time=400/3001.33 m s^-1 due east.