Question

A car initially at rest at point A starts to move with an acceleration of 3m/s2 for 5s and reaches point B. The car then applies brake and it stops at point C upon travelling a distance of 5m. From C the car starts to move at a uniform velocity of 5m/s for 1min and reaches point D. What is the distance between point A and D?

Answer 1

Answer:

Distance from A to D = 198.75m

Explanation:

Distance from A to B -

Given

$\blue\bigstar$Acceleration(a) = 3m/s²

$\blue\bigstar$Initial velocity (u) = 0m/s

$\blue\bigstar$Time taken (t) =5 seconds

Distance(s) -

• ${s=ut+\frac{1}{2}at² }$

• 1/2×3×5²
• 1/2×3×5
• 1/2×75
• Distance =37.5 m

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Distance from B to C - 5m

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Distance from C to D -

Given,

$\blue\bigstar$Initial velocity (u) = 0m/s

$\blue\bigstar$Final velocity (v) =5m/s

$\blue\bigstar$Time taken(t) = 1 min = 60 seconds

Acceleration(a) -

• ${a=\frac{v-u}{t}}$

• 5-0/60
• 5/60
• 0.083

Distance (s) -

• ${2as=v² - u² }$

• 2(0.83s)= 5²
• 0.16s= 25
• s= 25÷0.16
• Distance = 156.25 m

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Distance from A to D -

• Sum of all the given distances
• 156.25+5+37.5
• 198.75m