A train moves with a uniform velocity of 36km/h for 10 s. calculate the distance travelled.my answer's 100m. but my answer sheet states 6km. this is impossible because if you travel 6km in 10 secs, you'd cover 36km in 1 minute! you needn't take 1 hour. please check whether my answer is correct.
next, a car has a velocity of 10m/s. it now accelerates at 1m/s2 for 1/4min(15secs). find the distance travelled in this time and final speed of the car. since they're asking for this distance travelled during this acceleration period, then the car would have a straight line in a speed/time graph. to find distance, then we'd use the area under the speed/.time grpah to find distance. final speed is 25m/s so the distance can be found be 0.5 X 15 X 25. this gives me 187.5 m but the answer writes 262.5m.
the next question goes something like this
a tennis ball is hit vertically upwards with a velocity of 20m/s.
what is its deceleration as it moves upwards? is it correct to use the term 'deceleration'? i suppose so cuz objects accelerate at a constant speed when they fall without air resistance. but could someone clarify all these?
thanks!
1.
d = s*t
= 10m/s * 10s
= 100m
2.
d = area under graph
= area of rectangle + area of triangle
= 10m/s * 15s + 0.5 * 15m/s * 15s
= 262.5m
for qn 2, if u draw out correctly, the graph should consist of a rectangle and a triangle
3.
it is correct to use deceleration (which is -ve acceleration), which is the rate of decrease of velocity(?)
i dunno how to answer this qn >.>
For the first one, the answer is 10/3600 x 36 = .1km = 100m (you are correct)
2. s = (1/2)(u+v)(t)
= (1/2)(10+25)(15)
=1/2 * 35 * 15
= 262.5
Youll need to factor in the initial velocity too. If it starts at rest, it would be 187.5 but as it starts at 10ms-1, you would have to consider that when you work.
3. dunch noes. Ask ur teacher! I think that it is correct though. As it states that the upwards velocity is 20m/s, the upwards direction is postive.
Gah. someone beat me to the answers.
for the 3rd question. deceleration... that's gravity so if we take the positive y direction as "upwards" acceleration would be -9.81 m/s^2 or deceleration would be 9.81m/s^2 down.