Two cars $A$ and $B$ at rest at same point initially. If $A$ starts with uniform velocity of $40\, m/sec$ and $B$ starts in the same direction with constant acceleration of $4\,m/{s^2}$, then $B $ will catch $A$ after how much time.........$sec$
Two cars $A$ and $B$ at rest at same point initially. If $A$ starts with uniform velocity of $40\, m/sec$ and $B$ starts in the same direction with constant acceleration of $4\,m/{s^2}$, then $B $ will catch $A$ after how much time.........$sec$
$10$
$20$
$30$
$35$
Two cars $A$ and $B$ at rest at same point initially. If $A$ starts with uniform velocity of $40\, m/sec$ and $B$ starts in the same direction with constant acceleration of $4\,m/{s^2}$, then $B $ will catch $A$ after how much time.........$sec$
Let $A$ and $B$ will meet after time $t$ sec. it means the distance travelled by both will be equal.
${S_A} = ut = 40t$ and ${S_B} = \frac{1}{2}a{t^2} = \frac{1}{2} \times 4 \times {t^2}$
${S_A} = {S_B} \Rightarrow 40t = \frac{1}{2}4{t^2} \Rightarrow t = 20\,sec$
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