BigSaaks
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A cyclist heads off down a road at a steady speed of 8 m/s, and overtakes a stationary car.
Exactly 5 seconds after the bike overtakes, the car starts to accelerate at 4m/s per second uniformly, until it
reaches 72km/h, at which point it keeps the same steady speed.
a) Draw a velocity-time graph for the motion of each vehicle, on the same graph, from the point
where the bike overtakes, up until the point when the car catches up the bike.
b) Find the time that has elapsed between the two moments they are coincident
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RDKGames
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(Original post by BigSaaks)
A cyclist heads off down a road at a steady speed of 8 m/s, and overtakes a stationary car.
Exactly 5 seconds after the bike overtakes, the car starts to accelerate at 4m/s per second uniformly, until it
reaches 72km/h, at which point it keeps the same steady speed.
a) Draw a velocity-time graph for the motion of each vehicle, on the same graph, from the point
where the bike overtakes, up until the point when the car catches up the bike.
b) Find the time that has elapsed between the two moments they are coincident
I recommend you convert everything into metres for length and seconds for time, then graph the motion.
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Chakram
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(Original post by BigSaaks)
A cyclist heads off down a road at a steady speed of 8 m/s, and overtakes a stationary car.
Exactly 5 seconds after the bike overtakes, the car starts to accelerate at 4m/s per second uniformly, until it
reaches 72km/h, at which point it keeps the same steady speed.
a) Draw a velocity-time graph for the motion of each vehicle, on the same graph, from the point
where the bike overtakes, up until the point when the car catches up the bike.
b) Find the time that has elapsed between the two moments they are coincident
Hi there,

Im not too sure whether I'm right but for part B) I got 4seconds

The way I solved it is I created an equation for both the cyclist and the car using s=ut+1/2at^2
I then subbed one into another, removing s and then solved for t
You want to do this as the instant they meet their displacements will be the same.
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Physics Enemy
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(Original post by Chakram)
Hi there,

Im not too sure whether I'm right but for part B) I got 4seconds

The way I solved it is I created an equation for both the cyclist and the car using s=ut+1/2at^2
I then subbed one into another, removing s and then solved for t
You want to do this as the instant they meet their displacements will be the same.
Can't be right as car is at rest for 5s after bike passes. Then car takes 5s to accelerate from 0 to 20 m/s (72 km/h). They say maintaining this, so ans likely more than 10s. In the car eqn you used t instead of (t - 5).

10s after passing: car did 0.5*4*5^2 = 50 m, bike did 8*10 = 80 m (30 m ahead). Solve 20t = 30 + 8t then add 10s. Or 20(t - 10) = 30 + 8(t - 10) directly.
Last edited by Physics Enemy; 1 week ago
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