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) uniform velocity.

The slope of a position-time graph gives the velocity. A straight line has a constant slope, so the velocity is constant. A curve that gets steeper would show increasing velocity, and a line parallel to the time axis would show a body at rest.

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: (a) Both A and R are true and R is the correct explanation of A. If a body returns to its starting point, its initial and final positions are the same, so its displacement is zero, while the distance along the path it covered is not.

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.

A body starts with velocity $u$u and moves with uniform acceleration $a$a; after time $t$t its velocity is $v$v. Its velocity-time graph is a straight line from $(0,u)$(0,u) to $(t,v)$(t,v).

The slope of a velocity-time graph gives the acceleration:
$a=\text{slope}=\frac{\text{rise}}{\text{run}}=\frac{v-u}{t}.$a=slope=rise/run=(v-u)/t.
Multiplying by $t$t: $at=v-u$at=v-u, so
$v=u+at.$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=1/15≈0.067 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+1/2at²=0+1/2× 1/15×(300)²=90000/30=3000 m=3 km.

In one round the athlete covers the circumference $2\pi r$2π r in the time period $T$T, so
$\text{speed }v=\frac{\text{distance}}{\text{time}}=\frac{2\pi r}{T}.$speed v=distance/time=(2π r)/T.
$v=\frac{2\times\frac{22}{7}\times70}{88}=\frac{440}{88}=5\ \text{m s}^{-1}.$v=2× 22/7×7088=440/88=5 m s^-1.

The speed is constant, but the velocity is not: its direction keeps changing along the circle (it is along the tangent at each point). So uniform circular motion is an accelerated motion.

(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.
$a_1=\dfrac{20-0}{4}=5\ \text{m s}^{-2}.$a₁=(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.
$a_3=\dfrac{0-20}{2}=-10\ \text{m s}^{-2}\ (\text{retardation }10\ \text{m s}^{-2}).$a₃=(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=1/2(4+10)×20=1/2×14×20=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+1/2× t× at_triangle
$\boxed{s=ut+\tfrac12at^2}$s=ut+1/2at²

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+1/2×3×4²=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)=800/300≈2.67 m s^-1.

(iv) Average velocity $=\dfrac{\text{displacement}}{\text{total time}}=\dfrac{400}{300}\approx1.33\ \text{m s}^{-1}$=displacement/(total time)=400/300≈1.33 m s^-1 due east.