The acceleration is equal to (4-7)/t, where t is the amount of time on the rough patch. The length of the rough patch = 1/2(t)(7-4)=1.5t. The distance displaced after the rough patch is 4*2=8meters. The total diplacement =8+1.5t.
Rachel H.
asked • 01/13/21a speed skater moves across a frictionless ice at 7.0 m/s.
She slows steadily while on the rough patch, then continues on at 4m/s for 2s. a) What is her acceleration while on the rough patch? b) what's her total displacement?
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