Question

A body is moving with a velocity of 144 m/s. If it is a retired at a rate of 10 m/s², find the distance travelled by it before its velocity becomes 44 m/s.​

Answer 1

Given :-

Initial velocity = u = 144m/s

Final velocity = v = 44m/s

Acceleration = a = -10m/s² (since it retarded and i.e the speed decreased)

Distance = s = ?

Using 3rd equation of motion, that is 2as = v² - u²

➡ 2 × (-10) × s = (44)² - (144)²

➡ -20s = 1936 - 20736

➡ -20s = -18800

➡ s = -18800/-20

➡ s = 940m

Hence, the distance travelled by the body before it's velocity reached 44m/s is 940m.

Answer 2

Answer:

$\huge{\green{\sf{Distance(s)=940m}}}$

Explanation:

$\huge{\pink{\green{\underline{\red{\sf{SOLUTION-}}}}}}$

▪In this question information given about a moving body which initial velocity and final velocity is given and after retardation of 10m/s^2, final velocity is 44m/s.

▪So, we have to find the distance travelled during this retardation.

$\underline \bold{given : } \\ \to initial \: velocity(u) = 144m/s \\ \to final \: velocity(v) =44m/s \\ \to acceleration(a) = - 10m/ {s}^{2} \\ \\ \underline\bold{to \: find : } \\ \to Distance(s)= ?$

▪According to given question :

• We know the formula for 3rd eqn of motion.

▪So, putting all known values on that formula and only one unknown is their in formula that is distance.

$\to {v}^{2} - {u}^{2} = 2as \\ \to ({44})^{2} - ( {144})^{2} = 2 \times (- 10 )\times s \\ \to 1936 - 20736 = - 20 \times s \\ \to s = \frac{ - 18800}{ - 20} \\ \bold{\therefore s= 940m}$

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