An object is moving with constant acceleration. Its velocity is $48 m s^{- 1}$ at the end of $10$ second and becomes $68 m s^{- 1}$ at the end of the $15$ second. What would be the distance travelled by the object in $15$ second?

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Solution

Apply first kinematic equation of motion.
$v = u + a t$
$a = \frac{v - u}{t_{2} - t_{1}} = \frac{68 - 48}{15 - 10} = 4 m s^{- 2}$
Apply second kinematic equation of motion (from initial velocity, $u = 0$ )
$v^{2} - u^{2} = 2 a s$
$\Rightarrow 68^{2} - 0 = 2 \times 4 \times s$
$\Rightarrow s = 578 m$
Hence, distance travelled in $15 sec$ is $578 m$

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An object is moving with constant acceleration. Its velocity is 48m s−1 at the end of 10 second and becomes 68m s−1 at the end of the 15 second. What would be the distance travelled by the object in 15 second?

Apply first kinematic equation of motion.v=u+ata=v−ut2−t1=68−4815−10=4ms−2Apply second kinematic equation of motion (from initial velocity, u=0)v2−u2=2as⇒682−0=2×4×s⇒s=578mHence, distance travelled in 15sec is578m

An object is moving with constant acceleration. Its velocity is $48 m s^{- 1}$ at the end of $10$ second and becomes $68 m s^{- 1}$ at the end of the $15$ second. What would be the distance travelled by the object in $15$ second?