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(A) At $t = 5 \, s$,the velocity of the particle is $v_B = u + at = 0 + (1)(5) = 5 \, m/s$ and its position is $x_B = ut + \frac{1}{2}at^2 = 0 + \frac

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(A) At $t = 5 \, s$,the velocity of the particle is $v_B = u + at = 0 + (1)(5) = 5 \, m/s$ and its position is $x_B = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1)(5)^2 = 12.5 \, m$.Let the additional acceleration be $a'$. The total acceleration for $t > 5 \, s$ is $(1 + a')$.For the case with additional acceleration,at $t = 10 \, s$ (which is $5 \, s$ after the change):$v = v_B + (1 + a')(5) = 5 + 5 + 5a' = 10 + 5a'$$x = x_B + v_B(5) + \frac{1}{2}(1 + a')(5)^2 = 12.5 + 5(5) + 12.5(1 + a') = 12.5 + 25 + 12.5 + 12.5a' = 50 + 12.5a'$If the additional acceleration had not been provided,$a' = 0$:$v_0 = 5 + (1)(5) = 10 \, m/s$$x_0 = 12.5 + 5(5) + \frac{1}{2}(1)(5)^2 = 12.5 + 25 + 12.5 = 50 \, m$Given $x - x_0 = 12.5 \, m$:$(50 + 12.5a') - 50 = 12.5 \implies 12.5a' = 12.5 \implies a' = 1 \, m/s^2$.Then,$v - v_0 = (10 + 5a') - 10 = 5a' = 5(1) = 5 \, m/s$.

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