Question 137: Motion in 1D

Question

If the velocity of a particle is v = At + Bt2,  where A and B are constants, then the distance travelled by it between 1 s and 2 s is

Accepted Answer

Given,
v = At + Bt
2
$${{dx} \over {dt}}$$ =
At + Bt
2
$$\int {dx = \int {\left( {At + B{t^2}} ight)} } dt$$
x = $${{A{t^2}} \over 2} + {{B{t^3}} \over 3} + C$$
At t = 1, particle is at
x(t = 1) = $${A \over 2} + {B \over 3} + C$$
At t = 2, particle is at
x(t = 2) = $$2A + {{8B} \over 3} + C$$
$$ 	herefore $$ distance travelled by the particle between 1 s and 2 s is,
= x(t = 2) - x(t = 1)
= ($$2A + {{8B} \over 3} + C$$) - ($${A \over 2} + {B \over 3} + C$$)
= $${3 \over 2}A + {7 \over 3}B$$

Answer

$${3 \over 2}A + {7 \over 3}B$$

Answer

$${A \over 2} + {B \over 3}$$

Answer

$${3 \over 2}A + 4B$$

Answer

$$3A + 7B$$

Motion in 1D

If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is

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