Question

# A car starts from rest and moves along the x-axis with constant acceleration of $5m/{s}^{2}$ for $8seconds$. If it then continues with constant velocity, what distance will the car cover in $12seconds$ since it started from the rest?

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Solution

## Step 1: GivenInitial velocity, $u=0$Acceleration, $a=5m/{s}^{2}$Time, ${t}_{total}=12seconds\phantom{\rule{0ex}{0ex}}{t}_{1}=8seconds\phantom{\rule{0ex}{0ex}}{t}_{2}=4seconds$Step 2: Formula usedThe first equation of motion, $v=u+at$.The second equation of motion, ${s}_{1}=ut+\frac{1}{2}a{t}^{2}$.$Dis\mathrm{tan}ce=Speed×time$Step 3: Determine the distance covered in the first $8seconds$Let be the distance travelled in the first $8seconds$Using the second equation of motion, ${s}_{1}=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=u{t}_{1}+\frac{1}{2}a{{t}_{1}}^{2}\phantom{\rule{0ex}{0ex}}=0\left(8\right)+\frac{1}{2}\left(5\right){\left(8\right)}^{2}\phantom{\rule{0ex}{0ex}}=160m$${s}_{1}$Step 4: Determine the velocity at the end of 8 seconds when the car is acceleratingUsing the first equation of motion,$v=u+at\phantom{\rule{0ex}{0ex}}=u+a{t}_{1}\phantom{\rule{0ex}{0ex}}=0+\left(5\right)\left(8\right)\phantom{\rule{0ex}{0ex}}=40m/s$Step 5: Determine the remaining distanceThe distance traveled in first $8seconds$ has been calculated.The total time from the moment car starts from rest is $12seconds$Distance traveled in the last $4seconds$ will be ${s}_{2}$$Dis\mathrm{tan}ce=Speed×time\phantom{\rule{0ex}{0ex}}{s}_{2}=v×{t}_{2}\phantom{\rule{0ex}{0ex}}=40×4\phantom{\rule{0ex}{0ex}}=160m$Step 6: Determine the total distance$Totaldis\mathrm{tan}ce={s}_{1}+{s}_{2}\phantom{\rule{0ex}{0ex}}=160+160\phantom{\rule{0ex}{0ex}}=320m$Therefore, the car covers $320m$ in total.

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