At time t=0,a car moving along a straight lin | vedclass

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(D) Given $a = \frac{dv}{dt} = -0.5t$. Integrating with initial velocity $v(0) = 16 \; m/s$: $\int_{16}^{v} dv = \int_{0}^{t} -0.5t \; dt \implies v -

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All the above.

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(D) Given $a = \frac{dv}{dt} = -0.5t$. Integrating with initial velocity $v(0) = 16 \; m/s$: $\int_{16}^{v} dv = \int_{0}^{t} -0.5t \; dt \implies v - 16 = -0.25t^2 \implies v(t) = 16 - 0.25t^2$. The velocity changes direction when $v = 0$,so $16 - 0.25t^2 = 0 \implies t^2 = 64 \implies t = 8 \; s$. Thus,option $A$ is correct. Distance in $4 \; s$: $S_4 = \int_{0}^{4} (16 - 0.25t^2) dt = [16t - \frac{0.25t^3}{3}]_{0}^{4} = 64 - 5.33 = 58.67 \; m \approx 59 \; m$. Thus,option $B$ is correct. Distance in $10 \; s$: The particle moves forward until $t=8 \; s$ and backward from $t=8 \; s$ to $t=10 \; s$. Distance $d = \int_{0}^{8} v \; dt + |\int_{8}^{10} v \; dt|$. $\int_{0}^{8} (16 - 0.25t^2) dt = [16t - \frac{0.25t^3}{3}]_{0}^{8} = 128 - 42.67 = 85.33 \; m$. $\int_{8}^{10} (16 - 0.25t^2) dt = [16t - \frac{0.25t^3}{3}]_{8}^{10} = (160 - 83.33) - (128 - 42.67) = 76.67 - 85.33 = -8.66 \; m$. Total distance $= 85.33 + |-8.66| = 93.99 \; m \approx 94 \; m$. Thus,option $C$ is correct. Therefore,all statements are correct.

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 $a = -0.5t \; m/s^2$,where $t$ is in seconds. Mark the correct statement$(s)$.

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