Answer to Question #280561 in Physics for Kenny Bee

Question

A body at rest is given an initial uniform acceleration of 8.0m/s for 30s after which the acceleration is reduced to 5.0m/s for the next 20s. Draw the velocity-time graph of the motion using the information given above. Using the graph, calculate the (I) maximum speed attained during the motion. (ii) average retardation as the body is being brought to rest. (iii) total distance travelled during the first 50s. (iv) average speed during the same interval as in (ii).

Answer 1

(i) $v_{max}=a_1t_1=8\cdot30=240\ (m/s)$

(ii) $v=v_0-a_2t\to 0=240-5\cdot t\to t=48\ (s)$

$t=30+48=78\ (s)$ from the beginning of the movement.

(iii) $s_1=a_1t_1^2/2=8\cdot30^2/2=3600\ (m)$

$s_2=v_0t_2-a_2t_2^2/2=240\cdot20-5\cdot20^2/2=3800\ (m)$

$s_{total}=s_1+s_2=3600+3800=7400\ (m)$

(iv) $v_{avr}=\frac{s_1+s_2}{t_1+t_2}=\frac{3600+3800}{30+20}=148\ (m/s)$