Motion in One Dimension Class 9 ICSE Important Questions with Answers

Answer: (b) (speed, velocity).

Speed is a scalar (magnitude only) and velocity is a vector (magnitude and direction). Distance is scalar and displacement is vector.

Answer: (c) acceleration.

On a velocity-time graph, slope $=\dfrac{\Delta v}{\Delta t}=a$=(Δ v)/(Δ t)=a. The area under the graph gives the distance travelled.

Answer: (c) $10\text{ m s}^{-1}$10 m s^-1.

$v=u+at=0+(2)(5)=10\text{ m s}^{-1}$v=u+at=0+(2)(5)=10 m s^-1.

Answer: (b) $2\pi r$2π r and $0$0.

Distance is the full circumference $2\pi r$2π r. As the body returns to its start, the displacement (shortest distance between start and finish) is zero.

Answer: (a) If the body returns to its starting point, displacement $=0$=0 while distance (path length) is not zero; R correctly explains A.

Speed is the distance travelled per unit time; it is a scalar (has magnitude only) and is always positive.

Velocity is the displacement per unit time; it is a vector (has both magnitude and direction) and can be positive, negative or zero.

Given $u=20\text{ m s}^{-1}$u=20 m s^-1, $v=0$v=0, $t=5\text{ s}$t=5 s.

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

The acceleration is $-4\text{ m s}^{-2}$-4 m s^-2; the negative sign shows retardation (deceleration).

For uniform acceleration the velocity-time graph is a straight line rising from velocity $u$u at time $0$0 to velocity $v$v at time $t$t.

The slope of a velocity-time graph is the acceleration: $a=\dfrac{\text{change in velocity}}{\text{time}}=\dfrac{v-u}{t}$a=(change in velocity)/time=(v-u)/t.

So $at=v-u$at=v-u, which gives $v=u+at$v=u+at.

Convert: $v=72\text{ km h}^{-1}=72\times\dfrac{5}{18}=20\text{ m s}^{-1}$v=72 km h^-1=72× 5/18=20 m s^-1; $u=0$u=0; $t=2\text{ min}=120\text{ s}$t=2 min=120 s.

(i) $a=\dfrac{v-u}{t}=\dfrac{20-0}{120}=\boxed{0.167\text{ m s}^{-2}}$a=(v-u)/t=(20-0)/120=0.167 m s^-2

(ii) $s=\dfrac{u+v}{2}\times t=\dfrac{0+20}{2}\times120=\boxed{1200\text{ m}}$s=(u+v)/2× t=(0+20)/2×120=1200 m

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 top $v=0$v=0.

(i) $v^2=u^2+2as\Rightarrow 0=(19.6)^2+2(-9.8)s$v²=u²+2as⇒ 0=(19.6)²+2(-9.8)s
$s=\dfrac{(19.6)^2}{2\times9.8}=\dfrac{384.16}{19.6}=\boxed{19.6\text{ m}}$s=(19.6)²/2×9.8=384.16/19.6=19.6 m

(ii) $v=u+at\Rightarrow 0=19.6-9.8t$v=u+at⇒ 0=19.6-9.8t
$t=\dfrac{19.6}{9.8}=\boxed{2\text{ s}}$t=19.6/9.8=2 s

For uniform acceleration the velocity-time graph is a straight line from $u$u to $v$v in time $t$t. The distance $S$S is the area under it, a trapezium:
$S=\tfrac12(u+v)\,t.$S=1/2(u+v) t.
From the slope, $a=\dfrac{v-u}{t}$a=(v-u)/t, so $t=\dfrac{v-u}{a}$t=(v-u)/a. Substituting:
$S=\frac{(v+u)(v-u)}{2a}=\frac{v^2-u^2}{2a}\ \Rightarrow\ v^2=u^2+2aS.$S=(v+u)(v-u)/2a=(v²-u²)/2a ⇒ v²=u²+2aS.
Numerical: $v^2=10^2+2\times2\times75=100+300=400$v²=10²+2×2×75=100+300=400, so $v=20\ \text{m s}^{-1}$v=20 m s^-1.

The stone is dropped, so $u=0$u=0 and $a=g=10\text{ m s}^{-2}$a=g=10 m s^-2 (taking downward as positive).

(i) $S=ut+\tfrac12at^2\Rightarrow45=\tfrac12\times10\times t^2\Rightarrow t^2=9\Rightarrow t=\boxed{3\text{ s}}$S=ut+1/2at²⇒45=1/2×10× t²⇒ t²=9⇒ t=3 s.

(ii) $v=u+at=0+10\times3=\boxed{30\text{ m s}^{-1}}$v=u+at=0+10×3=30 m s^-1.

(iii) $S=\tfrac12\times10\times2^2=\boxed{20\text{ m}}$S=1/2×10×2²=20 m.

Derivation: For a body with initial velocity $u$u and uniform acceleration $a$a, the velocity-time graph is a straight line rising from $u$u to $v$v in time $t$t.

The distance $s$s equals the area under the graph, which is a trapezium = area of rectangle ($u\times t$u× t) + area of triangle ($\tfrac12\times t\times(v-u)$1/2× t×(v-u)).

Since $v-u=at$v-u=at,
$s=ut+\tfrac12\times t\times at=ut+\tfrac12at^2.$s=ut+1/2× t× at=ut+1/2at².

Numerical: $u=5\text{ m s}^{-1}$u=5 m s^-1, $a=2\text{ m s}^{-2}$a=2 m s^-2, $t=10\text{ s}$t=10 s.
$s=ut+\tfrac12at^2=5(10)+\tfrac12(2)(10)^2=50+100=\boxed{150\text{ m}}$s=ut+1/2at²=5(10)+1/2(2)(10)²=50+100=150 m

(i) The velocity-time graph rises from $0$0 to $20\text{ m s}^{-1}$20 m s^-1 (0-8 s), stays horizontal at $20\text{ m s}^{-1}$20 m s^-1 (8-20 s), then falls to $0$0 (20-25 s).

(ii) Acceleration $=\dfrac{20-0}{8}=\boxed{2.5\text{ m s}^{-2}}$=(20-0)/8=2.5 m s^-2. Retardation $=\dfrac{0-20}{5}=-4\text{ m s}^{-2}$=(0-20)/5=-4 m s^-2, i.e. $\boxed{4\text{ m s}^{-2}}$4 m s^-2.

(iii) Total distance = area under graph.
$s=\underbrace{\tfrac12(8)(20)}_{80}+\underbrace{(12)(20)}_{240}+\underbrace{\tfrac12(5)(20)}_{50}=\boxed{370\text{ m}}$s=1/2(8)(20)₈₀+(12)(20)₂₄₀+1/2(5)(20)₅₀=370 m

(i) Equal distances ($10\text{ m}$10 m) are covered in equal time intervals ($2\text{ s}$2 s), so the body is in uniform motion (constant speed).

(ii) $\text{speed}=\dfrac{\text{distance}}{\text{time}}=\dfrac{40}{8}=\boxed{5\text{ m s}^{-1}}$speed=distance/time=40/8=5 m s^-1.

(iii) The slope of a distance-time graph represents the speed of the body.

(iv) For a body at rest the distance does not change with time, so the distance-time graph is a straight horizontal line parallel to the time axis (zero slope).