Motion in One Dimension — ICSE Class 9 Physics Important Questions

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).

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) $v^2=u^2+2as\Rightarrow 0=(19.6)^2+2(-9.8)s$v^2=u^2+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

Distance in the $n^{\text{th}}$n^th second: $s_n=u+\dfrac{a}{2}(2n-1)$s_n=u+a/2(2n-1).

For $n=4$n=4: $u+\dfrac{a}{2}(7)=40\Rightarrow u+3.5a=40$u+a/2(7)=40 u+3.5a=40 ... (1)

For $n=6$n=6: $u+\dfrac{a}{2}(11)=60\Rightarrow u+5.5a=60$u+a/2(11)=60 u+5.5a=60 ... (2)

Subtracting (1) from (2): $2a=20\Rightarrow a=\boxed{10\text{ m s}^{-2}}$2a=20 a=10 m s^-2.

From (1): $u=40-3.5(10)=40-35=\boxed{5\text{ m s}^{-1}}$u=40-3.5(10)=40-35=5 m s^-1.

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)$12× t×(v-u)).

Since $v-u=at$v-u=at,
$s=ut+\tfrac12\times t\times at=ut+\tfrac12at^2.$s=ut+12× t× at=ut+12at^2.

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+12at^2=5(10)+12(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=12(8)(20)_80+(12)(20)_240+12(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).