Answer:
60 s
Step-by-step explanation:
Here is the complete question
Two students were racing home from school. Eric rides his skateboard and Robby rides his bicycle. Eric leaves 40 seconds before Robby. Eric traveled at a speed of 10 feet per second. Robby traveled at a speed of 30 feet per second. For how many seconds had Eric traveled when the both boys were the same distance from school?
Solution
Let Robby start his race at time t seconds. Eric starts his race at 40 seconds later implies his time is t' = t + 40.
Since Robby moves at a speed of 30 ft/s, the distance he travels in time, t seconds is d = 30t
Since Eric moves at a speed of 10 ft/s, the distance he travels in time, t' seconds is d' = 10t' = 10(t + 40).
When their distances are the same, d = d'. So,
30t = 10(t + 40)
expanding the bracket, we have,
30t = 10t + 400
collecting like terms, we have,
30t - 10t = 400
20t = 400
dividing both sides by 20, we have
t = 400/20
t = 20 s
Since the time Eric has travelled is t' = t + 40 = 20 + 40 = 60 s.
So, Eric has travelled 60 s when the boys have travelled the same distance from the school..
