Answer
displacement $160$ meters
distance $224$ meters
Work Step by Step
displacement
= $\int_0^{10}(36-24t+3t^{2})dt$
= $(36t-12t^{2}+t^{3})|_0^{10}$
= $160$ meters
distance
$36-24t+3t^{2}$ = $0$
$t^{2}-8t+12$ = $0$
$(t-6)(t-2)$ = $0$
$t$ = $2, 6$
= $\int_0^{2}(36-24t+3t^{2})dt$ + $\int_2^{6}(36-24t+3t^{2})dt$+ $\int_6^{10}(36-24t+3t^{2})dt$
= $(36t-12t^{2}+t^{3})|_0^{2}$ + $(36t-12t^{2}+t^{3})|_2^{6}$ + $(36t-12t^{2}+t^{3})|_6^{10}$
= $224$ meters
