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2.Motion in Straight Line
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At time $t=0$, a car moving along a straight line has a velocity of $16 \;m / s$. It slows down with an acceleration of $-0.5 t \;m / s ^2$, where $t$ is in second. Mark the correct statement (s).
AThe direction of velocity changes at $t=8\,s$
BThe distance travelled in $4\,s$ is approximately $59\,m$
CThe distance travelled by the particle in $10\,s$ is $94\,m$
DAll the above
Solution
(d)
$a=-0.5\,t$ or $\frac{d v}{d t}=-0.5\,t$
$\int \limits_{16}^v d v=\int \limits_0^t-(0.5 t) \cdot d t$
$v-16=-0.25 t^2$
$v=16-0.25 t^2$
Velocity will change its direction at the instant when $v=0$
$0=16-0.25 t^2$ or $t=8\,s$
$d_{4 s }=S_{4 s }=\int \limits_0^4 v d t$
$\int \limits_0^4\left(16-0.25 t^2\right) d t=58.66\,m =59\,m$
Now,
$S_{85}=\int \limits_0^8\left(16-0.25 t^2\right) d t=85.33\,m$
$S_{10 s }=\int \limits_0^{10}\left(16-0.25 t^2\right) d t=76.67\,m$
$d_{10 s } =A C+B C=A C+(A C-B A)$
$=2 A C-B A=94\,m$
$v_{10 s } =16-(0.25)(10)^2=-9\,m / s$
$a=-0.5\,t$ or $\frac{d v}{d t}=-0.5\,t$
$\int \limits_{16}^v d v=\int \limits_0^t-(0.5 t) \cdot d t$
$v-16=-0.25 t^2$
$v=16-0.25 t^2$
Velocity will change its direction at the instant when $v=0$
$0=16-0.25 t^2$ or $t=8\,s$
$d_{4 s }=S_{4 s }=\int \limits_0^4 v d t$
$\int \limits_0^4\left(16-0.25 t^2\right) d t=58.66\,m =59\,m$
Now,
$S_{85}=\int \limits_0^8\left(16-0.25 t^2\right) d t=85.33\,m$
$S_{10 s }=\int \limits_0^{10}\left(16-0.25 t^2\right) d t=76.67\,m$
$d_{10 s } =A C+B C=A C+(A C-B A)$
$=2 A C-B A=94\,m$
$v_{10 s } =16-(0.25)(10)^2=-9\,m / s$
Standard 11
Physics
Similar Questions
Match the following columns.
| colum $I$ | colum $II$ |
| $(A)$ Constant positive acceleration | $(p)$ Speed may increase |
| $(B)$ Constant negative acceleration | $(q)$ Speed may decrease |
| $(C)$ Constant displacement | $(r)$ Speed is zero |
| $(D)$ Constant slope of $a-t$ graph | $(s)$ Speed must increase |
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