36:T1149,

a = \frac{dv}{dt} = (2 \hat{i} + 6t \hat{j}) \, \text{ms}^{-2}

.2f:["$","li","1",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":2}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$36"}}]]}] 37:Tf32,At

t=1 \, \text{s}

, the acceleration

a

(2 \hat{i} + 6 \hat{j}) \, \text{ms}^{-2}

.30:["$","li","2",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":3}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$37"}}]]}] 38:Tb60,According to Newton's second law, the force

F

is equal to the mass

m

times acceleration

a

. The mass

m

is given as

500 \, \text{g}

, or equivalently,

0.5 \, \text{kg}

.31:["$","li","3",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":4}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$38"}}]]}] 39:T525,Therefore, the force

F

on the particle at

t=1 \, \text{s}

is:32:["$","li","4",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":5}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$39"}}]]}] 3a:T1474,

F = m \cdot a = 0.5 \cdot (2 \hat{i} + 6 \hat{j}) = (1 \hat{i} + 3 \hat{j}) \, \text{N}

.33:["$","li","5",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":6}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3a"}}]]}] 3b:Te64,So, the force acting on the particle at

t=1 \, \text{s}

(\hat{i} + x \hat{j}) \, \text{N}

, where

x=3

.34:["$","li","6",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":7}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3b"}}]]}] 35:["$","li","7",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":8}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"Therefore, the answer is

x=3

."}}]]}]

More on Laws of Motion

More on Laws of Motion