Question

A car travels with a constant velocity of 15m/s for 4 seconds and then decelerates to come to
halt in 4 seconds. Find the total displacement undergone by the car. Draw the v-t graph for the motion of this body.
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Answer 1

Solving

$\bull \: \large \sf \red{given : } \\ \\ \large \mapsto \rm \: velocity = 15m {s}^{ - 1} \\ \\ \large \rm \mapsto \: time = 4 \: seconds$

time of decelerate = 4 seconds

$\large \purple{ \sf \: formula : }$

Decelerate =

$\large \sf \leadsto \: \dfrac{initial \: \: velocity - final \: \: velocity}{time \: \: taken}$

Initial Velocity = u

Final Velocity = v

Time = t

Since,

$\large \looparrowright\rm \blue{velocity \: \: is \: \: constant \: ..}$

Suppose , car started from rest

$\large \rm\implies \: initial \: \: velocity = 0m {s}^{ - 1}$

★ Final Velocity = 15m/s

★ Time taken = 4 seconds

$\:$

$\large \sf \green{ \rightarrow \: finding \: \: decelerate : }$

$\large \tt : \implies \: \dfrac{0 - 15}{4} \\ \\ \\ \large \tt : \implies \: \dfrac{ - 15}{4} = - 3.75$

$\large \rm \orange{finding \: \: distance : }$

$\large \mapsto \sf \: s = ut + \dfrac{1}{2} a {t}^{2} \\ \\ \\ \large \implies \sf \: s = 0(4) + \frac{ - 3.75}{2} \times {4}^{2} \\ \\ \\ \large \implies \sf \: s = \frac{ - 375}{ 100 \times 2} \times {4} \times 4 \\ \\ \\ \large \implies \sf \: s = \frac{ - 375}{ 100 \times \cancel2} \times \cancel{4} \: \: 2 \times 4$

$\large \rm \therefore \: \red{distance = - 30m {s}^{ - 1} }$

Displacement = - 30m/s